🧠:Alright, let's see. I need to prove that ( 45 < x_{1000} < 45.1 ) given the recursive sequence ( x_0 = 5 ) and ( x_{n+1} = x_n + frac{1}{x_n} ). Hmm, okay. So the sequence is defined by starting at 5 and each subsequent term is the previous term plus the reciprocal of the previous term. I need to figure out how this sequence behaves over 1000 steps and show that it ends up between 45 and 45.1. First, let me try to get an intuitive sense of what's happening here. Each term is increasing because we're adding a positive number (( frac{1}{x_n} )) each time. So the sequence is strictly increasing. That means ( x_n ) grows without bound as ( n ) approaches infinity, right? But how fast does it grow? Since each step adds ( frac{1}{x_n} ), which gets smaller as ( x_n ) gets larger, the increments are decreasing. So the growth rate slows down as ( n ) increases.But to get precise bounds for ( x_{1000} ), I need a more rigorous approach. Maybe I can model this recursion with a differential equation? If I think of ( x_{n+1} - x_n = frac{1}{x_n} ), this resembles a difference equation. If ( n ) is large and the step size is small, perhaps I can approximate it as a differential equation. Let me try that.Let’s denote ( x_n ) as a function ( x(n) ). Then the difference ( x(n+1) - x(n) approx frac{dx}{dn} ). So, the equation becomes:( frac{dx}{dn} = frac{1}{x} )This is a separable differential equation. Let's separate variables:( x , dx = dn )Integrating both sides:( frac{1}{2}x^2 = n + C )Solving for ( x ):( x = sqrt{2n + C} )Now, apply the initial condition. When ( n = 0 ), ( x = 5 ):( 5 = sqrt{2(0) + C} implies C = 25 )So the approximate solution is:( x(n) approx sqrt{2n + 25} )Hmm, if this approximation holds, then ( x_{1000} approx sqrt{2*1000 + 25} = sqrt{2025} = 45 ). But the problem states that ( x_{1000} ) is greater than 45 but less than 45.1. So the approximation gives exactly 45, but the actual value is slightly higher. That suggests that the differential equation approximation is a bit off, probably because it's a continuous approximation of a discrete process. The actual sequence might grow a little faster than the continuous model predicts because each step's increment is based on the current value, not an averaged rate. So maybe the integral approximation underestimates the true value?Alternatively, let's analyze the difference between the actual sequence and the approximation. Let me formalize this. Let’s denote ( x_n^2 ). Then:( x_{n+1}^2 = left( x_n + frac{1}{x_n} right)^2 = x_n^2 + 2 + frac{1}{x_n^2} )Therefore, the recursion for ( x_n^2 ) is:( x_{n+1}^2 = x_n^2 + 2 + frac{1}{x_n^2} )This seems useful. If I can express this recursion in terms of ( x_n^2 ), maybe I can telescope the sum. Let's write:( x_{n+1}^2 - x_n^2 = 2 + frac{1}{x_n^2} )Then, summing from ( n = 0 ) to ( N-1 ):( x_N^2 - x_0^2 = sum_{n=0}^{N-1} left( 2 + frac{1}{x_n^2} right) = 2N + sum_{n=0}^{N-1} frac{1}{x_n^2} )Given ( x_0 = 5 ), so ( x_0^2 = 25 ). Therefore:( x_N^2 = 25 + 2N + sum_{n=0}^{N-1} frac{1}{x_n^2} )Therefore, the exact value of ( x_N^2 ) is ( 25 + 2N + S ), where ( S = sum_{n=0}^{N-1} frac{1}{x_n^2} ). So for ( N = 1000 ):( x_{1000}^2 = 25 + 2000 + sum_{n=0}^{999} frac{1}{x_n^2} = 2025 + S )Therefore, ( x_{1000} = sqrt{2025 + S} ). Since all terms in ( S ) are positive, ( S > 0 ), so ( x_{1000} > sqrt{2025} = 45 ). That gives the lower bound. Now, we need to show that ( S < 0.1 times 90.1 ) (wait, no). Wait, ( x_{1000} < 45.1 ), so ( x_{1000}^2 < (45.1)^2 ). Let me compute ( 45.1^2 ):( 45^2 = 2025 ), so ( 45.1^2 = (45 + 0.1)^2 = 45^2 + 2*45*0.1 + 0.1^2 = 2025 + 9 + 0.01 = 2034.01 )Therefore, to have ( x_{1000}^2 < 2034.01 ), we need ( 2025 + S < 2034.01 ), which implies ( S < 9.01 ). So we need to show that ( sum_{n=0}^{999} frac{1}{x_n^2} < 9.01 ). Therefore, if we can bound the sum ( S = sum_{n=0}^{999} frac{1}{x_n^2} ) above by 9.01, then we have the upper bound. So the problem reduces to estimating this sum.Given that ( x_n ) is increasing, ( frac{1}{x_n^2} ) is decreasing. Therefore, the terms of the sum are decreasing. So maybe we can approximate the sum by an integral or use a comparison with a series.Alternatively, note that ( x_n^2 = 25 + 2n + S_n ), where ( S_n = sum_{k=0}^{n-1} frac{1}{x_k^2} ). Wait, that seems recursive. Let me see. For each n, ( x_n^2 = 25 + 2n + sum_{k=0}^{n-1} frac{1}{x_k^2} ). So, ( S_n = sum_{k=0}^{n-1} frac{1}{x_k^2} = x_n^2 - 25 - 2n ). Therefore, for each n, ( frac{1}{x_n^2} = frac{1}{25 + 2n + S_n} ). But ( S_n ) is positive, so ( frac{1}{x_n^2} < frac{1}{25 + 2n} ). Therefore, ( S = sum_{n=0}^{999} frac{1}{x_n^2} < sum_{n=0}^{999} frac{1}{25 + 2n} ).But wait, the sum ( sum_{n=0}^{999} frac{1}{25 + 2n} ) can be approximated as an integral. Let's consider the general term ( frac{1}{25 + 2n} ). Let’s make a substitution: let m = n + 12.5, then 25 + 2n = 2(n + 12.5) = 2m. So,( sum_{n=0}^{999} frac{1}{25 + 2n} = frac{1}{2} sum_{n=0}^{999} frac{1}{n + 12.5} )This sum is similar to the harmonic series. The harmonic series ( H_N = sum_{k=1}^N frac{1}{k} ) is approximately ( ln N + gamma ), where ( gamma ) is Euler-Mascheroni constant (~0.5772). So, the sum ( sum_{k=1}^{N} frac{1}{k + c} ) can be approximated as ( ln(N + c) - ln(1 + c) ). In our case, the sum starts from n=0 to 999, which is 1000 terms. So,( sum_{n=0}^{999} frac{1}{n + 12.5} = sum_{m=12.5}^{1011.5} frac{1}{m} approx ln(1011.5) - ln(12.5) )Compute this:First, approximate ( ln(1011.5) ). Since ( ln(1000) = 6.9078 ), and ( ln(1011.5) ≈ ln(1000) + ln(1.0115) ≈ 6.9078 + 0.0114 ≈ 6.9192 ).Then, ( ln(12.5) ≈ ln(12) + ln(1.0417) ≈ 2.4849 + 0.0408 ≈ 2.5257 ).So, the difference is approximately ( 6.9192 - 2.5257 ≈ 4.3935 ).Therefore, the original sum is approximately ( frac{1}{2} * 4.3935 ≈ 2.19675 ).Wait, that's only about 2.2, which is much less than 9.01. But this seems conflicting with our earlier conclusion that ( S < sum_{n=0}^{999} frac{1}{25 + 2n} approx 2.2 ), which would imply ( x_{1000}^2 ≈ 2025 + 2.2 ≈ 2027.2 ), so ( x_{1000} ≈ sqrt{2027.2} ≈ 45.02 ), which is still within 45.1. But the problem states to prove ( x_{1000} < 45.1 ), but according to this rough approximation, it's even less. However, my initial calculation using the differential equation gave 45, and the sum S was around 2.2, leading to 45.02. But the problem says 45 < x_{1000} < 45.1, so this seems feasible, but why does the upper bound need to be 45.1? Maybe my integral approximation is too rough.Wait, but if S is approximately 2.2, then x_{1000} is about sqrt(2025 + 2.2) ≈ sqrt(2027.2) ≈ 45.02. So maybe the actual sum S is a bit larger, leading to x_{1000} around 45.05, which is still less than 45.1. So perhaps the upper bound is not tight, but the problem just wants us to show it's less than 45.1. However, how do we know S isn't larger?Alternatively, perhaps my approach is wrong here. Wait, in the step where I said ( x_{n}^2 = 25 + 2n + S_n ), where ( S_n = sum_{k=0}^{n-1} frac{1}{x_k^2} ). Therefore, S_n is part of x_n^2. But when I bounded S_n by the harmonic series, I assumed ( S_n < sum frac{1}{25 + 2k} ), but in reality, since each x_k^2 = 25 + 2k + S_k, which is greater than 25 + 2k, so 1/x_k^2 < 1/(25 + 2k). Therefore, the sum S_n = sum_{k=0}^{n-1} 1/x_k^2 < sum_{k=0}^{n-1} 1/(25 + 2k). So indeed, the upper bound on S_n is less than the sum of reciprocals of 25 + 2k.But when I approximated that sum as approximately 2.2, which would lead to x_{1000} ≈ 45.02, which is within the required bound. So perhaps the problem is expecting us to use this method to bound S, and thus show that x_{1000}^2 = 2025 + S_{1000} < 2025 + 9.01, but maybe we need a better bound on S.Alternatively, maybe my integral approximation is missing something. Let me recast the sum ( sum_{n=0}^{999} frac{1}{25 + 2n} ). Let me write it as ( frac{1}{2} sum_{n=0}^{999} frac{1}{n + 12.5} ). This is equivalent to ( frac{1}{2} left( sum_{n=12.5}^{1011.5} frac{1}{n} right) ). But the integral from 12.5 to 1011.5 of 1/x dx is ln(1011.5) - ln(12.5) ≈ 6.919 - 2.526 ≈ 4.393 as before. So the sum is approximately equal to the integral plus some error term. The integral is a lower bound for the sum if we use the left endpoint, but actually, since the function 1/x is decreasing, the sum ( sum_{k=m}^{n} frac{1}{k} ) is less than the integral from m to n+1 of 1/x dx. Wait, no. Let's recall that for decreasing functions, the sum from m to n of f(k) is less than the integral from m to n+1 of f(x) dx.Wait, maybe it's better to use the integral test for series. For a decreasing function f(x), the sum from k = m to n of f(k) ≤ integral from m-1 to n of f(x) dx. Wait, I need to recall the exact inequality.Actually, for a decreasing function, ( int_{a}^{b} f(x) dx leq sum_{k=a}^{b-1} f(k) leq int_{a}^{b} f(x) dx + f(a) ). Maybe. Let me think. Suppose f(x) is decreasing. Then, for each interval [k, k+1], f(k+1) ≤ integral_{k}^{k+1} f(x) dx ≤ f(k). So summing from k = m to n-1:sum_{k=m}^{n-1} f(k+1) ≤ integral_{m}^{n} f(x) dx ≤ sum_{k=m}^{n-1} f(k)Which rearranges to:sum_{k=m+1}^{n} f(k) ≤ integral_{m}^{n} f(x) dx ≤ sum_{k=m}^{n-1} f(k)Hence, integral_{m}^{n} f(x) dx ≤ sum_{k=m}^{n-1} f(k) ≤ integral_{m}^{n} f(x) dx + f(m)But in our case, the sum is from n=0 to 999 of 1/(25 + 2n). Let’s write it as sum_{k=12.5}^{1011.5} 1/(2k) with substitution. Wait, maybe not. Let me consider the substitution m = n + 12.5, then the sum becomes sum_{m=12.5}^{1011.5} 1/(2m). But integrating 1/(2m) from 12.5 to 1011.5 gives (1/2)(ln(1011.5) - ln(12.5)) ≈ (1/2)(4.393) ≈ 2.196, as before. Then, the sum is approximately equal to this integral. But how much error is there?Since the function 1/(2m) is decreasing, the sum from m=12.5 to 1011.5 (but m takes integer steps?) Wait, no. The original substitution was m = n + 12.5, but n is integer, so m is 12.5, 13.5, ..., 1011.5. So the sum is over non-integer m, which complicates the integral approximation. Alternatively, maybe consider shifting the index.Alternatively, let me approximate sum_{n=0}^{999} 1/(25 + 2n). Let’s write this as sum_{k=25}^{2024} step 2 of 1/k. Wait, no. Wait, when n=0, 25 + 2n =25; when n=1, 25 +2=27; n=2, 29;... up to n=999: 25 + 2*999=25 +1998=2023. So the sum is 1/25 +1/27 +1/29 +...+1/2023. So it's the sum of reciprocals of odd numbers starting from 25 up to 2023, stepping by 2 each time.Alternatively, note that sum_{k=25}^{2023} frac{1}{k} where k is odd. But how does this relate to the harmonic series? The sum of reciprocals of odd numbers up to N is (1/2)(H_{(N-1)/2} + ln 2)). Wait, maybe not. Let me recall that the sum of reciprocals of all numbers up to N is H_N, and the sum of reciprocals of even numbers up to N is (1/2)H_{N/2}, and the sum of reciprocals of odd numbers up to N is H_N - (1/2)H_{N/2}. But maybe this is more complicated.Alternatively, approximate the sum by integrating 1/x from 25 to 2023, but accounting for step size 2. Since each term is 2 apart, the integral approximation would be (1/2) * ln(2023/25). Let me compute that.First, compute the integral from 25 to 2023 of (1/x) dx = ln(2023) - ln(25) ≈ ln(2023/25) ≈ ln(80.92) ≈ 4.394. Then, since the step is 2, the sum is approximately (1/2)*4.394 ≈ 2.197. So this matches the previous approximation. Therefore, the sum is approximately 2.197. However, since we are stepping through every other term, the actual sum is less than the integral approximation? Wait, for decreasing functions, using the left endpoint overestimates the integral. Wait, actually, when you approximate the sum with step size 2, each term is the value at the left endpoint, so the sum overestimates the integral. Therefore, the integral from 25 to 2023 of 1/x dx ≈ 4.394, so the sum is ≈ (1/2)(integral + correction). Hmm, maybe this is getting too convoluted.Alternatively, let's use the integral test to bound the sum. For a decreasing function f(x), the sum from n = a to b of f(n) ≤ integral_{a-1}^{b} f(x) dx. Wait, maybe better to use:For an decreasing function f,integral_{a}^{b+1} f(x) dx ≤ sum_{n=a}^{b} f(n) ≤ integral_{a-1}^{b} f(x) dxBut in our case, the function is f(n) = 1/(25 + 2n). Let me adjust variables for clarity. Let’s set m = 25 + 2n. Then, when n=0, m=25; when n=999, m=25 + 1998 = 2023. So the sum is sum_{m=25}^{2023 step 2} 1/m. But how to bound this sum? Let's consider that sum_{m=25}^{2023 step 2} 1/m = (1/2) sum_{m=25}^{2023} 1/m where m is odd. But the sum of 1/m over odd m from 25 to 2023 can be written as (sum_{m=1}^{2023} 1/m - sum_{m=1}^{24} 1/m) - (sum_{m=2}^{2022 step 2} 1/m - sum_{m=2}^{24 step 2} 1/m). Wait, this seems complicated. Alternatively, approximate the sum of reciprocals of odd numbers as half the harmonic series. But actually, the sum of reciprocals of all numbers up to N is H_N, and the sum of reciprocals of even numbers up to N is (1/2)H_{N/2}, so the sum of reciprocals of odd numbers up to N is H_N - (1/2)H_{N/2}. But for large N, H_N ≈ ln N + γ, so the sum of reciprocals of odd numbers up to N is approximately ln N + γ - (1/2)(ln(N/2) + γ) = ln N + γ - (1/2)(ln N - ln 2 + γ) = (1/2) ln N + (1/2)γ + (1/2) ln 2. Therefore, sum_{m=1}^{M odd} 1/m ≈ (1/2)(ln M + γ + ln 2). But in our case, the sum is from m=25 to 2023 odd. So it's equal to sum_{m=1}^{2023 odd} 1/m - sum_{m=1}^{23 odd} 1/m. Using the approximation:sum_{m=1}^{2023 odd} 1/m ≈ (1/2)(ln 2023 + γ + ln 2)sum_{m=1}^{23 odd} 1/m ≈ (1/2)(ln 23 + γ + ln 2)Therefore, the difference is approximately (1/2)(ln 2023 - ln 23) = (1/2) ln(2023/23) ≈ (1/2) ln(88) ≈ (1/2)(4.477) ≈ 2.2385Therefore, the sum from m=25 to 2023 odd is approximately 2.2385. Then, since our original sum is half of that (since step is 2), but wait, no. Wait, if the sum of reciprocals of odd numbers up to 2023 is approximately 2.2385, then the sum from 25 to 2023 odd would be approximately 2.2385 - sum_{m=1}^{23 odd} 1/m. Wait, but we already subtracted sum up to 23. Maybe this is getting too fuzzy. Alternatively, perhaps the approximation gives us around 2.2, which is similar to the integral estimate. So regardless of the method, the sum S is around 2.2, leading to x_{1000}^2 ≈ 2025 + 2.2 = 2027.2, so x_{1000} ≈ 45.02. Therefore, 45 < x_{1000} < 45.1 holds.But the problem statement wants a proof, so we need rigorous bounds, not just approximations. Let's think about how to bound S more carefully.We have ( x_{n}^2 = 25 + 2n + sum_{k=0}^{n-1} frac{1}{x_k^2} ). Therefore, each term ( x_n^2 > 25 + 2n ). Hence, ( frac{1}{x_n^2} < frac{1}{25 + 2n} ). Therefore, the sum ( S = sum_{n=0}^{999} frac{1}{x_n^2} < sum_{n=0}^{999} frac{1}{25 + 2n} ).So if we can bound this sum above by 9.01, then we can get the desired upper bound. But from our previous approximations, the sum is around 2.2, which is way less than 9.01. Therefore, the upper bound is easily satisfied, so x_{1000}^2 = 2025 + S < 2025 + 9.01 = 2034.01, hence x_{1000} < 45.1.But wait, why does the problem specify 45.1 as the upper bound if our estimate is much lower? Maybe the approximations are too rough, and the actual sum S could be higher? Let's check with a smaller n.For example, let's compute the first few terms manually. Let's compute x_0 to x_3 and see how the sum S behaves.x_0 = 5x_1 = 5 + 1/5 = 5.2x_2 = 5.2 + 1/5.2 ≈ 5.2 + 0.1923 ≈ 5.3923x_3 ≈ 5.3923 + 1/5.3923 ≈ 5.3923 + 0.1853 ≈ 5.5776Compute S_3 = 1/x_0² + 1/x_1² + 1/x_2² = 1/25 + 1/(5.2)^2 + 1/(5.3923)^2Calculate each term:1/25 = 0.041/(5.2)^2 ≈ 1/27.04 ≈ 0.0371/(5.3923)^2 ≈ 1/29.08 ≈ 0.0344So S_3 ≈ 0.04 + 0.037 + 0.0344 ≈ 0.1114Now, according to our approximation using the integral, the sum up to n=3 would be sum_{k=0}^3 1/(25 + 2k) = 1/25 + 1/27 + 1/29 + 1/31 ≈ 0.04 + 0.037 + 0.0345 + 0.0323 ≈ 0.1438. Comparing with the actual S_3 ≈ 0.1114, which is less than the sum of 1/(25 + 2k). Therefore, our initial upper bound is valid, since S_n < sum_{k=0}^{n-1} 1/(25 + 2k). So for n=3, actual S is 0.1114 vs upper bound 0.1438, so the upper bound is a bit loose. Therefore, the sum S_n is less than the sum of 1/(25 + 2k), which is itself less than something. Wait, but we have already established that S_n < sum_{k=0}^{n-1} 1/(25 + 2k), and we can bound that sum. If we can show that sum_{k=0}^{999} 1/(25 + 2k) < 9.01, then x_{1000} < 45.1. But according to our integral approximation, the sum is about 2.2, which is way less than 9.01, so the bound is easily satisfied.Therefore, even if our upper bound is loose, it's still sufficient to prove the required inequality. Therefore, the problem's upper limit of 45.1 is quite generous, and our analysis shows that x_{1000} is approximately 45.02, which is within the required bounds.But perhaps I need to formalize this into a proof. Let me structure the steps:1. Show that ( x_n ) is increasing and thus ( x_{n} geq 5 ) for all n, so ( frac{1}{x_n} leq frac{1}{5} ).2. Derive the recursion for ( x_n^2 ):( x_{n+1}^2 = x_n^2 + 2 + frac{1}{x_n^2} ).3. Telescope the recursion to get:( x_N^2 = 25 + 2N + sum_{n=0}^{N-1} frac{1}{x_n^2} ).4. Let ( S_N = sum_{n=0}^{N-1} frac{1}{x_n^2} ). Then ( x_N^2 = 25 + 2N + S_N ).5. Since ( x_n geq 5 ), ( frac{1}{x_n^2} leq frac{1}{25} ), so ( S_N leq frac{N}{25} ). However, for N=1000, this would give ( S_{1000} leq 40 ), which is too large. So this bound is not useful.6. Instead, observe that ( x_n^2 = 25 + 2n + S_n geq 25 + 2n ). Therefore, ( frac{1}{x_n^2} leq frac{1}{25 + 2n} ).7. Hence, ( S_N = sum_{n=0}^{N-1} frac{1}{x_n^2} leq sum_{n=0}^{N-1} frac{1}{25 + 2n} ).8. Compute the sum ( sum_{n=0}^{N-1} frac{1}{25 + 2n} ). For N=1000, this is sum_{k=25}^{2023 step 2} 1/k. As we approximated before, this sum is approximately 2.2. However, to bound it above, we can use the integral test.9. The sum ( sum_{n=0}^{999} frac{1}{25 + 2n} ) can be compared to the integral of ( frac{1}{25 + 2x} ) from x = -0.5 to x = 999.5 to account for overestimation.Wait, another way. Since the function ( f(n) = frac{1}{25 + 2n} ) is decreasing, the sum from n=0 to 999 is less than the integral from n=-0.5 to n=999.5 of f(n) dn.Compute the integral:( int_{-0.5}^{999.5} frac{1}{25 + 2n} dn )Let u = 25 + 2n, then du = 2 dn, dn = du/2. When n = -0.5, u = 25 + 2*(-0.5) = 24. When n = 999.5, u = 25 + 2*999.5 = 25 + 1999 = 2024.Thus, the integral becomes:( frac{1}{2} int_{24}^{2024} frac{1}{u} du = frac{1}{2} (ln 2024 - ln 24) ≈ frac{1}{2} (7.613 - 3.178) ≈ frac{1}{2} (4.435) ≈ 2.2175 )Therefore, the sum ( sum_{n=0}^{999} frac{1}{25 + 2n} < 2.2175 + f(0) ), but actually, the integral from -0.5 to 999.5 gives a lower bound for the sum? Wait, no. For a decreasing function, the sum from n=a to n=b is less than the integral from a-1 to b of f(x) dx. Wait, perhaps better to use:For decreasing f, ( sum_{n=m}^{k} f(n) leq int_{m-1}^{k} f(x) dx ).So, applying this:( sum_{n=0}^{999} f(n) leq int_{-1}^{999} f(x) dx ).But f(x) = 1/(25 + 2x). Then,( int_{-1}^{999} frac{1}{25 + 2x} dx ).Let u = 25 + 2x, du = 2 dx, limits from u=25 + 2*(-1)=23 to u=25 + 2*999=2023.Thus,( frac{1}{2} int_{23}^{2023} frac{1}{u} du = frac{1}{2} (ln 2023 - ln 23) ≈ frac{1}{2} (7.613 - 3.135) ≈ frac{1}{2} (4.478) ≈ 2.239 ).Therefore, the sum is less than approximately 2.24. Hence, ( S_{1000} < 2.24 ), so ( x_{1000}^2 = 2025 + S_{1000} < 2025 + 2.24 = 2027.24 ), so ( x_{1000} < sqrt{2027.24} ≈ 45.02 ), which is indeed less than 45.1. Therefore, the upper bound holds.To get a more precise upper bound, perhaps we can calculate the exact sum using the integral plus the first term. Wait, but even if we take the integral approximation and add the first term, we get 2.24 + 1/25 ≈ 2.24 + 0.04 = 2.28, but actually, the integral already accounted for the entire sum. Hmm, no, the integral from -1 to 999 is used to bound the sum from 0 to 999. Therefore, the sum is less than 2.24, as per the integral.Therefore, combining all this:Since ( x_{1000}^2 = 2025 + S_{1000} ), and ( S_{1000} < 2.24 ), we have ( x_{1000}^2 < 2027.24 ), so ( x_{1000} < sqrt{2027.24} ≈ 45.02 ), which is less than 45.1. Therefore, the upper bound holds.For the lower bound, since ( S_{1000} > 0 ), ( x_{1000}^2 > 2025 ), so ( x_{1000} > 45 ).Therefore, combining both results, ( 45 < x_{1000} < 45.1 ). But wait, the problem states to prove ( x_{1000} < 45.1 ), which we have shown via bounding S_{1000} < 2.24 leading to x_{1000} < 45.02, which is tighter. But maybe the problem just wants us to show the much easier bound of 45.1, which is well within our calculated upper limit.Alternatively, perhaps the problem expects a different approach, such as using induction or more refined estimates. Let me think.Another approach: consider the difference between ( x_{n+1}^2 ) and ( x_n^2 ). As established before:( x_{n+1}^2 - x_n^2 = 2 + frac{1}{x_n^2} ).Summing from n=0 to N-1:( x_N^2 - x_0^2 = 2N + sum_{n=0}^{N-1} frac{1}{x_n^2} ).Thus,( x_N^2 = 25 + 2N + sum_{n=0}^{N-1} frac{1}{x_n^2} ).Let’s denote ( S_N = sum_{n=0}^{N-1} frac{1}{x_n^2} ). Then,( S_N = x_N^2 - 25 - 2N ).But also, since ( x_{n}^2 geq 25 + 2n ), each term ( frac{1}{x_n^2} leq frac{1}{25 + 2n} ).Therefore,( S_N leq sum_{n=0}^{N-1} frac{1}{25 + 2n} ).We already bounded this sum above by approximately 2.24 for N=1000, which gives ( x_{1000}^2 < 2025 + 2.24 ), hence ( x_{1000} < 45.02 ).But perhaps to get a better lower bound for S, we can note that ( x_n^2 = 25 + 2n + S_n ), so each term ( frac{1}{x_n^2} = frac{1}{25 + 2n + S_n} geq frac{1}{25 + 2n + S_N} ), assuming S_n is increasing. Wait, but since S_n is the sum up to n, S_n increases with n. Therefore, ( frac{1}{x_n^2} geq frac{1}{25 + 2n + S_{1000}} ) for all n < 1000.Therefore, the sum ( S_{1000} = sum_{n=0}^{999} frac{1}{x_n^2} geq sum_{n=0}^{999} frac{1}{25 + 2n + S_{1000}} ).But this leads to an inequality:( S_{1000} geq sum_{n=0}^{999} frac{1}{25 + 2n + S_{1000}} ).Let’s denote ( T = S_{1000} ), then:( T geq sum_{n=0}^{999} frac{1}{25 + 2n + T} ).This seems difficult to solve directly, but perhaps we can approximate. Let's approximate the sum on the right:( sum_{n=0}^{999} frac{1}{25 + 2n + T} ).Again, substitute m = n + (25 + T)/2. Wait, but 25 + 2n + T = 2(n + (25 + T)/2). So, similar to before, this sum can be approximated by an integral:( sum_{n=0}^{999} frac{1}{25 + 2n + T} ≈ frac{1}{2} lnleft( frac{25 + 2*999 + T}{25 + T} right) ).But 25 + 2*999 + T = 2023 + T, and 25 + T is the lower limit. So,( frac{1}{2} lnleft( frac{2023 + T}{25 + T} right) ).Therefore, the inequality becomes:( T geq frac{1}{2} lnleft( frac{2023 + T}{25 + T} right) ).This is a transcendental equation in T. Let's plug in T ≈ 2.2 as before:Left side: 2.2Right side: 0.5 * ln( (2023 + 2.2)/(25 + 2.2) ) ≈ 0.5 * ln(2025.2 / 27.2) ≈ 0.5 * ln(74.44) ≈ 0.5 * 4.31 ≈ 2.155So 2.2 ≥ 2.155, which holds. If we try T=2.155:Left side: 2.155Right side: 0.5 * ln(2025.155 /27.155) ≈ 0.5 * ln(74.44) ≈ 2.155, which matches. So this suggests that T ≈ 2.155 is the fixed point. Therefore, our initial approximation of T≈2.2 is reasonable, and this gives a better lower bound for T.But this seems a bit circular. However, combining both upper and lower bounds, we can sandwich T between approximately 2.15 and 2.24, leading to x_{1000} between sqrt(2025 + 2.15) ≈ 45.01 and sqrt(2025 + 2.24) ≈ 45.02. Therefore, x_{1000} is approximately 45.015, safely within the required bounds.But since the problem only requires proving 45 < x_{1000} <45.1, and our analysis shows it's around 45.015, the bounds are satisfied. Therefore, the key steps are:1. Derive the squared recursion and telescope to relate x_{1000}^2 to the sum S.2. Show S is positive, giving the lower bound.3. Bound S above by comparing to an integral, proving the upper bound.Therefore, the final answer is that ( 45 < x_{1000} <45.1 ), as required.

✅:To prove that ( 45 < x_{1000} < 45.1 ) for the sequence ( x_0 = 5 ) and ( x_{n+1} = x_n + frac{1}{x_n} ), we analyze the behavior of ( x_n^2 ):1. Recursion for ( x_n^2 ): [ x_{n+1}^2 = left(x_n + frac{1}{x_n}right)^2 = x_n^2 + 2 + frac{1}{x_n^2}. ] Telescoping this from ( n = 0 ) to ( N-1 ): [ x_N^2 = x_0^2 + 2N + sum_{n=0}^{N-1} frac{1}{x_n^2}. ] For ( N = 1000 ), this becomes: [ x_{1000}^2 = 25 + 2000 + sum_{n=0}^{999} frac{1}{x_n^2} = 2025 + S_{1000}, ] where ( S_{1000} = sum_{n=0}^{999} frac{1}{x_n^2} ).2. Bounding ( S_{1000} ): - Lower Bound: Since ( S_{1000} > 0 ), we have ( x_{1000}^2 > 2025 ), so ( x_{1000} > 45 ). - Upper Bound: Each term ( frac{1}{x_n^2} leq frac{1}{25 + 2n} ). Thus: [ S_{1000} leq sum_{n=0}^{999} frac{1}{25 + 2n}. ] Approximating this sum using integrals: [ sum_{n=0}^{999} frac{1}{25 + 2n} approx frac{1}{2} lnleft(frac{2023}{25}right) approx 2.24. ] Therefore, ( S_{1000} < 2.24 ), leading to: [ x_{1000}^2 < 2025 + 2.24 = 2027.24 implies x_{1000} < sqrt{2027.24} approx 45.02. ]Since ( 45 < x_{1000} < 45.02 ), the bounds ( 45 < x_{1000} < 45.1 ) hold.Final Answer:[boxed{45 < x_{1000} < 45.1}]

🔑:We aim to prove the inequalities ( 45 < x_{1000} < 45.1 ) given the recursive sequence ( x_{n+1} = x_n + frac{1}{x_n} ) with ( x_0 = 5 ).To achieve this, we will show that for all ( n in mathbb{N} ), the following inequalities hold:[ sqrt{2n + 25} leq x_n leq sqrt{2n + 25} + 0.1. ]By evaluating these at ( n = 1000 ), we can derive the desired bounds for ( x_{1000} ).# Step 1: Induction HypothesisAssume for induction that:[ sqrt{2k + 25} leq x_k leq sqrt{2k + 25} + 0.1 ]for some ( k ). # Step 2: Establish the Recurrent RelationFirst, note that the recurrence relation can be rewritten as:[ x_{k+1} = x_k + frac{1}{x_k}. ]# Step 3: Lower Bound Use the inequality ( x_k^2 leq x_{k+1}^2 ) to establish the lower bound for ( x_k ). Start by considering the form:[ x_{k+1}^2 = x_k^2 + 2 left(frac{1}{x_k}right) x_k + left(frac{1}{x_k}right)^2 = x_k^2 + 2 + frac{1}{x_k^2}. ]Since ( frac{1}{x_k^2} > 0 ) for any ( x_k ), we can simplify this to:[ x_{k+1}^2 > x_k^2 + 2. ]Summing up these inequalities from ( k = 0 ) to ( n-1 ), we get:[ x_n^2 > x_0^2 + 2n. ]Given ( x_0 = 5 ):[ x_n^2 > 25 + 2n. ]Taking square roots, we obtain the lower bound:[ x_n > sqrt{2n + 25}. ]# Step 4: Upper BoundTo establish the upper bound, consider the assumption ( x_k < sqrt{2k + 25} + 0.1 ). We need to show that this implies:[ x_{k+1} < sqrt{2(k+1) + 25} + 0.1. ]Using the recurrence relation again, we have:[ x_{k+1} = x_k + frac{1}{x_k}. ]Given ( x_k geq 5 ), ( frac{1}{x_k} leq 0.2 ). Therefore:[ x_{k+1} < x_k + 0.2. ]Suppose ( x_k leq sqrt{2k + 25} + 0.1 ), then:[ x_{k+1} leq sqrt{2k + 25} + 0.3. ]Now consider:[ sqrt{2(k+1) + 25} = sqrt{2k + 27}. ]Note that ( sqrt{2k + 27} leq sqrt{2k + 25} + 0.1 ) for large ( n ). Thus:[ x_{k+1} leq sqrt{2k + 25} + 0.3 leq sqrt{2(k+1) + 25} + 0.1. ]# Step 5: ConclusionThus, by induction, for all ( n ):[ sqrt{2n + 25} leq x_n leq sqrt{2n + 25} + 0.1. ]Specifically, for ( n = 1000 ):[ sqrt{2 cdot 1000 + 25} leq x_{1000} leq sqrt{2 cdot 1000 + 25} + 0.1. ][ sqrt{2025} leq x_{1000} leq sqrt{2025} + 0.1. ]Since ( sqrt{2025} = 45 ):[ 45 leq x_{1000} leq 45.1. ]Therefore, we proved the required bounds:[ 45 < x_{1000} < 45.1. ][boxed{45 < x_{1000} < 45.1}]

🧠:Okay, so I need to figure out the ratio in which the plane passing through points B, C, and D divides the diagonal AA' of a cuboid. Let me start by visualizing a cuboid. A cuboid has 8 vertices, right? Let's say the cuboid is oriented with vertex A at the origin, and edges along the x, y, and z axes. Then, the other vertices can be labeled accordingly. So, vertex A is at (0, 0, 0). Then, moving along the edges, vertex B would be along the x-axis at (a, 0, 0), vertex D along the y-axis at (0, b, 0), and vertex C along the z-axis at (0, 0, c). Wait, no, actually, in a standard cuboid, the edges from A would be AB, AD, and AE? Hmm, maybe I need to clarify the labeling. Maybe the cuboid is labeled such that from vertex A, the adjacent vertices are B, D, and E, with B along x, D along y, and E along z. Then, the opposite vertex A' would be at (a, b, c). But the problem mentions points B, C, D as the endpoints of the edges originating from A. Wait, that seems conflicting because if edges originate from A, then the endpoints should be B, D, and another vertex. Maybe the labeling is different here. Let me check the problem statement again.The problem says: "the plane passing through the endpoints B, C, D of the edges originating from vertex A." So, from vertex A, three edges go out to B, C, and D. That means A is connected to B, C, and D. Then, the cuboid must have edges AB, AC, and AD. But in a standard cuboid, there are only three edges from each vertex, so maybe in this case, the cuboid is not axis-aligned? Or perhaps the labeling is different. Wait, perhaps the cuboid is such that A is connected to B, C, D, which are adjacent vertices. Then, the other vertices can be constructed by moving along the edges. Let me try to assign coordinates.Let's assume vertex A is at (0, 0, 0). Then, if edges AB, AC, AD are along the three axes, let's assign coordinates as follows:- A: (0, 0, 0)- B: (l, 0, 0) along the x-axis- C: (0, m, 0) along the y-axis- D: (0, 0, n) along the z-axisWait, but in that case, the cuboid would have vertices at (l, m, n), but the edges from A would be AB, AC, AD. Then, the opposite vertex A' would be at (l, m, n). The diagonal AA' goes from (0,0,0) to (l, m, n). The plane passing through B, C, D is the plane that contains the points (l,0,0), (0,m,0), and (0,0,n). So, I need to find where this plane intersects the diagonal AA', and then determine the ratio in which this intersection divides AA'.First, I need to find the equation of the plane passing through B, C, D. Let's denote the coordinates as follows:- B: (l, 0, 0)- C: (0, m, 0)- D: (0, 0, n)Wait, but hold on. If B, C, D are endpoints of edges from A, then depending on the cuboid's edges, these points could be along different axes. But in a standard cuboid, each vertex is connected via three edges along the x, y, z axes. So, if A is at (0,0,0), then the adjacent vertices would be (l,0,0), (0,m,0), (0,0,n). Then, the opposite vertex A' would be (l, m, n). The diagonal AA' is the line from (0,0,0) to (l, m, n). The plane passing through B, C, D is the plane containing (l,0,0), (0,m,0), (0,0,n).So, I need to find the equation of this plane. To do that, let's use the general equation of a plane: ax + by + cz = d. Since the plane passes through B, C, D, we can substitute their coordinates into the equation.For point B (l, 0, 0): a*l + b*0 + c*0 = d => a*l = d.For point C (0, m, 0): a*0 + b*m + c*0 = d => b*m = d.For point D (0, 0, n): a*0 + b*0 + c*n = d => c*n = d.Therefore, from these three equations, we have:a = d/lb = d/mc = d/nTherefore, the equation of the plane can be written as (d/l)x + (d/m)y + (d/n)z = d. Dividing both sides by d (assuming d ≠ 0, which it must be since otherwise the plane equation would be trivial and not pass through those points), we get:x/l + y/m + z/n = 1.So the equation of the plane is x/l + y/m + z/n = 1.Now, the diagonal AA' is the line from (0,0,0) to (l, m, n). We can parametrize this line. Let's let t be a parameter such that when t=0, we are at A (0,0,0), and when t=1, we are at A' (l, m, n). So the parametric equations are:x = l*ty = m*tz = n*tWe need to find the value of t where this line intersects the plane x/l + y/m + z/n = 1.Substituting the parametric equations into the plane equation:(l*t)/l + (m*t)/m + (n*t)/n = 1Simplifying:t + t + t = 1 => 3t = 1 => t = 1/3.Therefore, the point of intersection is at t = 1/3, so the coordinates are (l/3, m/3, n/3).Therefore, the plane divides the diagonal AA' into two parts: from A (0,0,0) to (l/3, m/3, n/3), and from (l/3, m/3, n/3) to A' (l, m, n).The ratio of the lengths of these two segments is the ratio of t to (1 - t), which is 1/3 to 2/3, so the ratio is 1:2.Wait, but hold on. Let me verify that. Since the parameter t is 1/3, the distance from A to the intersection point is t times the length of AA', and the distance from the intersection point to A' is (1 - t) times the length of AA'. Therefore, the ratio is t : (1 - t) = 1/3 : 2/3 = 1:2.Therefore, the plane divides the diagonal AA' in the ratio 1:2, with the segment closer to A being 1 part and the segment closer to A' being 2 parts.But wait, let me confirm this with another approach to be safe. Let's compute the vector equation of the diagonal AA' and find the intersection with the plane.The diagonal AA' can be represented as the line (0,0,0) + t(l, m, n), t ∈ [0,1]. The plane equation is x/l + y/m + z/n = 1.Plugging in x = l*t, y = m*t, z = n*t into the plane equation gives:(l*t)/l + (m*t)/m + (n*t)/n = t + t + t = 3t = 1 => t = 1/3. So same result.Therefore, the intersection is at t = 1/3, hence ratio 1:2.Alternatively, maybe using mass point geometry or similar triangles?But I think the parametric approach is straightforward here. So the plane cuts the diagonal at 1/3 of the way from A to A', so the ratio is 1:2.Wait, but let me consider if the cuboid has edges of different lengths, does this ratio depend on l, m, n? The answer here seems to be no, because in the equation substitution, the l, m, n canceled out, leading to t = 1/3 regardless of their values. Therefore, regardless of the cuboid's dimensions, the ratio is always 1:2.Is that possible? Let's take a simple case where the cuboid is a cube, so l = m = n. Then, the plane passing through B, C, D would be the plane x + y + z = l. The diagonal is from (0,0,0) to (l,l,l). The intersection point would be at (l/3, l/3, l/3), so t = 1/3. So ratio 1:2. Makes sense.Another test case: take a cuboid with l=1, m=2, n=3. The plane equation is x/1 + y/2 + z/3 = 1. The diagonal is from (0,0,0) to (1,2,3). Parametrize as (t, 2t, 3t). Plug into plane equation: t/1 + (2t)/2 + (3t)/3 = t + t + t = 3t = 1 => t=1/3. So intersection at (1/3, 2/3, 1). Then, the distance from A is sqrt( (1/3)^2 + (2/3)^2 + 1^2 ) and from A' is sqrt( (2/3)^2 + (4/3)^2 + 2^2 ). Wait, but actually, since we are dealing with ratios along the diagonal, which is a straight line, the ratio is determined by the parameter t. So even if the cuboid is not a cube, the ratio remains 1:2.Therefore, regardless of the cuboid's dimensions, the plane through B, C, D divides the diagonal AA' in the ratio 1:2.Hence, the answer should be 1:2.Final AnswerThe plane divides the diagonal ( AA' ) in the ratio boxed{1:2}.

✅:To determine the ratio in which the plane passing through the endpoints ( B, C, D ) of the edges originating from vertex ( A ) divides the diagonal ( AA' ) of a cuboid, we start by assigning coordinates to the vertices. Let vertex ( A ) be at ( (0, 0, 0) ), and the adjacent vertices ( B, C, D ) be at ( (l, 0, 0) ), ( (0, m, 0) ), and ( (0, 0, n) ) respectively. The opposite vertex ( A' ) is at ( (l, m, n) ).The equation of the plane passing through points ( B, C, D ) can be derived using the general plane equation ( ax + by + cz = d ). Substituting the coordinates of ( B, C, D ) into this equation, we find:- For ( B(l, 0, 0) ): ( al = d )- For ( C(0, m, 0) ): ( bm = d )- For ( D(0, 0, n) ): ( cn = d )Solving these, we get ( a = frac{d}{l} ), ( b = frac{d}{m} ), and ( c = frac{d}{n} ). The plane equation simplifies to:[frac{x}{l} + frac{y}{m} + frac{z}{n} = 1]The diagonal ( AA' ) is parameterized as ( (lt, mt, nt) ) where ( t ) ranges from 0 to 1. Substituting this into the plane equation:[frac{lt}{l} + frac{mt}{m} + frac{nt}{n} = 1 implies t + t + t = 1 implies 3t = 1 implies t = frac{1}{3}]Thus, the intersection point divides ( AA' ) at ( t = frac{1}{3} ). The ratio of the segments from ( A ) to the intersection and from the intersection to ( A' ) is ( frac{1}{3} : frac{2}{3} = 1:2 ).Therefore, the plane divides the diagonal ( AA' ) in the ratio (boxed{1:2}).

🔑:Let's denote the points of the cuboid as follows: (A, B, C, D) for the vertices originating from point (A) and the centrally symmetric vertices as (A', B', C', D'), respectively (as seen in the figure).1. Identify Important Diagonals: - The diagonal (A A') is the space diagonal of the cuboid. - The diagonal (A D) is a face diagonal, and it is intersected by another face diagonal (B C) at point (F). The intersection splits (B C) in half.2. Understand the Symmetry in the Problem: - The plane intersecting the points (B), (C), and (D) intersects the space diagonal (A A') at point (T). We need to determine the ratio in which (T) divides (A A').3. Surface Diagonal and Triangle Similarity: - The diagonals intersecting in the plane within the face of the cuboid provide the necessary geometrical ratios. - Given the similarity of the triangles formed by the intersections, we can analyze the respective segments.4. Use Triangle Similarity: - Consider the triangles formed: - ( triangle AFT ) and ( triangle A'DT ) are similar. - Since these triangles are similar, corresponding sides are proportional: [ frac{AT}{TA'} = frac{AF}{A'D} ]5. Calculate the Ratios: - ( AF ) is the half-diagonal of the face ( ABD ), and ( A'D ) is twice this length. - Therefore: [ frac{AF}{A'D} = frac{AF}{2AF} = frac{1}{2} ] - This implies that (A T : T A') is in the ratio (1:2).6. Conclusion: - Thus, the plane through (B, C), and (D) intersects the space diagonal (A A') such that the segment (A T) is one-third the length of (A A'), slicing off one-third of the space diagonal length.[boxed{text{The plane divides the space diagonal in the ratio } 1:2.}]

🧠:Okay, so I need to figure out the greatest possible number of chords that can have at least one common point with a line passing through point A₁ on a circle with 2017 distinct points. All possible chords are drawn between these points, and the line through A₁ doesn't pass through any other points A₂ to A₂₀₁₇. Hmm, let me think through this step by step.First, let's visualize the problem. There's a circle with 2017 points labeled A₁ to A₂₀₁₇. All possible chords between these points are drawn. Then, there's a line that starts at A₁ and doesn't go through any of the other points. I need to find the maximum number of chords that this line can intersect. So, the line can intersect chords either at A₁ or somewhere else along their length. But since the line passes through A₁, all chords starting at A₁ will definitely intersect the line at A₁. Then, the question is also about how many other chords (not starting at A₁) can this line intersect somewhere else on the circle.Wait, but the line can't pass through any other points A₂ to A₂₀₁₇, so it can only intersect chords that cross it somewhere in between their endpoints. So, chords that have one endpoint on one side of the line and the other endpoint on the other side of the line will be intersected by the line. But since the line passes through A₁, the position of the line relative to the circle will divide the remaining 2016 points into two arcs: one on each side of the line. The number of chords that cross the line would then be the number of chords with one endpoint in each arc.But the line can be rotated around A₁ to maximize this number. Since the points are fixed on the circle, rotating the line will change which points are on each side. To maximize the number of chords crossing the line, we need to arrange the line such that the number of points on each side is as balanced as possible. Because the number of chords between two sets is maximized when the sets are as equal as possible. For example, if there are n points on one side and m on the other, the number of chords between them is n*m. To maximize n*m with n + m = 2016 (since A₁ is on the line, not counted in either side), the product is maximized when n and m are as close as possible. Since 2016 is even, n and m would be 1008 each. Then, the number of chords crossing the line would be 1008*1008. But wait, is that correct?But also, we have to remember that chords from A₁ to all other points are already intersected at A₁. So those chords are 2016 in total. Then, the chords not involving A₁ that cross the line would be 1008*1008. So total chords intersected would be 2016 + 1008*1008. Wait, but 1008*1008 is 1,016,064. Adding 2016 gives 1,018,080. But the problem is asking for the greatest possible number of chords that can have at least one common point with the line. So that includes both the chords through A₁ and the chords crossing the line elsewhere. But is that correct?Wait, but chords that start at A₁ and go to another point are only intersected at A₁. The line passes through A₁, so all these chords share the point A₁ with the line. Then, the other chords that cross the line somewhere else on the circle (not at another point) are those that have one endpoint on each side of the line. So the total number of intersected chords would be the 2016 chords from A₁ plus the number of chords crossing the line, which is 1008*1008. Therefore, the total is 2016 + (1008)^2. Let me compute that.But wait, 1008*1008 is 1008 squared. Let me compute 1000 squared is 1,000,000, plus 8 squared is 64, plus 2*1000*8*2 = 32,000. Wait, no, (a + b)^2 = a² + 2ab + b². So 1008² = (1000 + 8)² = 1000² + 2*1000*8 + 8² = 1,000,000 + 16,000 + 64 = 1,016,064. Then adding 2016, which is 2,016, gives 1,016,064 + 2,016 = 1,018,080.But let me check if that's the maximum. If the line is arranged so that it splits the remaining points into two equal halves, 1008 on each side, then the number of chords crossing the line is 1008*1008. If the line is not balanced, say 1007 and 1009, then the product would be 1007*1009 = (1008 -1)(1008 +1) = 1008² -1 = 1,016,064 -1 = 1,016,063, which is less. So indeed, balancing gives the maximum. So the maximum number of chords crossed by the line is 1,018,080. But wait, let me confirm again. The problem states "at least one common point with this line". So chords that pass through A₁ (i.e., chords connecting A₁ to any other point) are counted once each, and chords that cross the line somewhere else are also counted. So the total is indeed the sum of these two quantities.But is this correct? Let's take a smaller example to verify. Suppose there are 3 points on a circle: A₁, A₂, A₃. Then all chords are A₁A₂, A₁A₃, A₂A₃. If we draw a line through A₁ that doesn't pass through A₂ or A₃. Depending on the position of the line, it can split the other two points. If the line is such that A₂ and A₃ are on the same side, then the chord A₂A₃ is not crossed by the line. If the line is between them, splitting them, then chord A₂A₃ is crossed. But in this case, with 3 points, the maximum number of chords intersected would be 2 (A₁A₂, A₁A₃) plus 1 (A₂A₃) if the line splits them. But since there are only two other points, the line can only split them if it's placed between them. So the total chords would be 3. But let's compute using the formula: n = 3, so remaining points after A₁ is 2. Splitting into 1 and 1. Then chords crossing the line would be 1*1 =1. Then chords from A₁ are 2. Total 3, which matches. So the formula works here.Another example: 5 points. A₁ and four others. If the line splits the four others into two groups of 2 and 2. Then chords crossing the line would be 2*2=4. Chords from A₁:4. Total 8. Let's see, all chords from A₁ (4) plus chords between the two groups (4). But total chords in the circle are C(5,2)=10. So if the line intersects 8 chords, that seems correct. If the line is arranged to split the other four points into 3 and 1, then chords crossing would be 3*1=3. So total 4+3=7, which is less. So the formula works here.Therefore, returning to the original problem. 2017 points. The line passes through A₁ and splits the remaining 2016 points into two equal groups of 1008 each. Then chords crossing the line are 1008*1008, and chords from A₁ are 2016. So total is 2016 + 1008². Let's compute 1008² again: 1008*1008. As I did earlier, that's 1,016,064. Adding 2016: 1,016,064 + 2,016 = 1,018,080. So that's the total number.But wait, the problem says "the greatest possible number of chords that can have at least one common point with this line". So, this would include all chords that either have A₁ as an endpoint or cross the line somewhere else. So yes, that seems right.But let me check again. Suppose I have a chord that connects A₁ to some point. The line passes through A₁, so that chord shares A₁ with the line. Then, any other chord that connects two points, one on each side of the line, will cross the line. Therefore, those chords are intersected by the line. So yes, those two sets (chords from A₁ and chords crossing the line) are distinct. A chord can't be in both sets unless it's from A₁ and crosses the line again, but since chords are straight lines, a chord from A₁ can't cross the line again unless the line is tangent, but since all chords are between two points on the circle, the line through A₁ can only intersect a chord from A₁ at A₁ itself. So no other intersection. Therefore, the two sets are disjoint.Therefore, the total number of chords intersected is indeed the sum of the two quantities: 2016 + 1008². Therefore, the answer is 2016 + 1008² = 2016 + 1,016,064 = 1,018,080. But let me check if 1008 is indeed half of 2016. 2016 divided by 2 is 1008, yes. So splitting 2016 points into two groups of 1008 each maximizes the product. Therefore, the number of crossing chords is 1008*1008.Therefore, the final answer should be 1,018,080. But let me check if there's any possible way to get more. Suppose the line passes through A₁ and is arranged such that it's tangent to the circle. Wait, but the points are on the circle, so a tangent line at A₁ would only touch the circle at A₁, but the line in the problem is passing through A₁ and not passing through any other points. So it's a secant line passing through A₁, intersecting the circle at A₁ and another point, but since all other points are marked, the line can't pass through any other marked points. Therefore, the line is a secant that only passes through A₁, and then extends to infinity on both sides. Therefore, the line divides the circle into two arcs, each containing some number of points. The maximum number of chords crossing the line is when the number of points on each arc is as equal as possible.Wait, but actually, the line is straight, so passing through A₁ and another point would pass through two points, but in our case, the line doesn't pass through any other points. Therefore, the line is near A₁ but doesn't pass through any other points, effectively dividing the circle into two arcs with 1008 points on each side. Therefore, the product is maximized.Therefore, I think the answer is 2016 + 1008*1008 = 1,018,080. But let me check with another approach.Alternatively, the total number of chords that intersect the line is equal to the number of chords with one endpoint on each side of the line plus the number of chords that have A₁ as an endpoint. Since the line passes through A₁, all chords from A₁ are included. For the other chords, they must cross the line, which requires one endpoint on each side. So if the line divides the remaining 2016 points into two sets of size k and 2016 - k, then the number of crossing chords is k*(2016 - k). To maximize k*(2016 - k), since 2016 is even, maximum occurs at k = 1008, giving 1008². Therefore, total is 2016 + 1008² = 1,018,080. Therefore, the answer is 1,018,080.But the problem states "at least one common point with this line". So chords that have A₁ as an endpoint have A₁ in common with the line, and chords that cross the line have a different intersection point. Therefore, all those chords are counted. Therefore, the answer is indeed 2016 + 1008² = 1,018,080. But wait, let me check if this is correct for another case. For example, n=5 points. A₁, and 4 others. Then, the maximum number of chords would be 4 (from A₁) + 2*2=4, total 8. Which is correct. For n=5, total chords is 10. So 8 chords intersected. If the line is arranged to split the 4 points into 2 and 2. So that seems right.Therefore, the formula works for smaller numbers, so likely works for 2017. Therefore, the answer is 1008² + 2016 = 1,016,064 + 2,016 = 1,018,080. So boxed, that would be boxed{1018080}.Final Answerboxed{1018080}

✅:To find the greatest possible number of chords that can have at least one common point with a line passing through point (A_1) on a circle with 2017 distinct points, we need to consider two types of chords: those that start at (A_1) and those that cross the line elsewhere.1. Chords starting at (A_1): There are 2016 such chords, as (A_1) can be connected to each of the other 2016 points.2. Chords crossing the line: The line through (A_1) divides the remaining 2016 points into two arcs. To maximize the number of chords crossing the line, the points should be divided as evenly as possible. Since 2016 is even, the optimal division is 1008 points on each side. The number of chords crossing the line is then the product of the number of points on each side, which is (1008 times 1008).Adding these two results together gives the total number of chords intersected by the line:[2016 + 1008^2]Calculating the values:[1008^2 = 1016064][2016 + 1016064 = 1018080]Thus, the greatest possible number of chords that can have at least one common point with the line is (boxed{1018080}).

🔑:To find the maximum number of chords on a circle that intersect a specified line through point ( A_1 ) and not passing through any other points ( A_2, ldots, A_{2017} ), we proceed as follows:1. Divide the points based on the given line: Let the line through ( A_1 ) divide the remaining 2016 points into two sets. Denote the number of points on one side of the line by ( k ). Therefore, the number of points on the other side is ( 2016 - k ).2. Count the chords that intersect the line: Each chord that intersects the line must connect a point on one side of the line to a point on the other side. Hence, the number of such chords is given by the product ( k(2016 - k) ).3. Count the chords through ( A_1 ): Since ( A_1 ) is fixed and connects to every other point independently of the division, there are always ( 2016 ) chords passing through ( A_1 ).4. Expressing the total intersection count: The total number of chords that either intersect the line (i.e., connecting points from opposite sides) or pass through ( A_1 ) is thus: [ k(2016 - k) + 2016 ]5. Maximizing ( k(2016 - k) ): We now find the value of ( k ) that maximizes ( k(2016 - k) ). Rewrite the quadratic expression: [ k(2016 - k) = 2016k - k^2 ] This is a downward-opening parabola of the form ( -k^2 + 2016k ). The maximum value of such a quadratic function occurs at: [ k = frac{2016}{2} = 1008 ]6. Calculating at ( k = 1008 ): Substitute ( k = 1008 ) in the quadratic term: [ 1008(2016 - 1008) = 1008 times 1008 = 1008^2 = 1008^2. ]7. Add the fixed number of chords through ( A_1 ): Including the 2016 fixed chords: [ k(2016 - k) + 2016 = 1008^2 + 2016 ]8. Final computation: Calculate ( 1008^2 ): [ 1008^2 = 1008 times 1008 = 1016064 ] Then, add 2016: [ 1016064 + 2016 = 1018080 ] Conclusion[boxed{1018080}]

🧠:Okay, so I need to prove that if for real numbers a, b, c, the inequality (a + b + c)c < 0 holds, then the discriminant of the quadratic equation ax² + bx + c = 0 is positive, which is b² - 4ac > 0. Hmm, let's start by understanding what each part of the problem means.First, the given condition is (a + b + c)c < 0. Let me parse this. The product of (a + b + c) and c is negative. That means one of them is positive and the other is negative. So either (a + b + c) is positive and c is negative, or (a + b + c) is negative and c is positive.Now, the discriminant of a quadratic equation ax² + bx + c = 0 is indeed b² - 4ac. If the discriminant is positive, the quadratic has two distinct real roots. So the problem is linking the condition (a + b + c)c < 0 to the quadratic having two real roots. Maybe there's a connection here with evaluating the quadratic at certain points?Wait, if I plug x = 1 into the quadratic, I get a(1)² + b(1) + c = a + b + c. So (a + b + c) is the value of the quadratic at x = 1. Similarly, c is the value of the quadratic at x = 0, because plugging x = 0 gives a(0) + b(0) + c = c. So the given condition (a + b + c)c < 0 is saying that the product of the quadratic evaluated at x = 1 and x = 0 is negative. That is, f(1) * f(0) < 0, where f(x) = ax² + bx + c.Now, if the product of f(1) and f(0) is negative, that means one of them is positive and the other is negative. By the Intermediate Value Theorem, if a continuous function changes sign between two points, it must cross the x-axis somewhere between those points. Since f(x) is a quadratic function, which is continuous, then between x = 0 and x = 1, there must be at least one real root. But wait, a quadratic can have at most two real roots. If there's a root between 0 and 1, does that guarantee the discriminant is positive?Wait, if there's at least one real root, then the discriminant is non-negative (i.e., ≥ 0). But the problem states that the discriminant is strictly greater than 0. So how do we go from knowing there's a root between 0 and 1 to the discriminant being positive? Maybe because quadratics can have two real roots, and if one is between 0 and 1, perhaps the other is somewhere else, making the discriminant positive. But maybe there's another way.Alternatively, maybe we can use the given inequality directly to derive the discriminant. Let's consider that (a + b + c)c < 0. Let's denote S = a + b + c. Then the inequality is S * c < 0. So S and c have opposite signs. So either S > 0 and c < 0, or S < 0 and c > 0.Let me write down the discriminant: D = b² - 4ac. I need to show that D > 0. Let's see if we can relate S and c to D.Express S as a + b + c. Let's try to express a in terms of S, b, and c: a = S - b - c. Substitute this into the discriminant:D = b² - 4ac = b² - 4(S - b - c)c = b² - 4Sc + 4bc + 4c².Hmm, not sure if that's helpful. Maybe there's another way. Let's consider cases based on the sign of c.Case 1: c > 0. Then, from the given inequality S * c < 0, since c is positive, S must be negative. So a + b + c < 0.Case 2: c < 0. Then, S must be positive, so a + b + c > 0.Maybe I can analyze both cases.Case 1: c > 0 and a + b + c < 0.Case 2: c < 0 and a + b + c > 0.Let's take Case 1 first. c > 0, and a + b + c < 0. Let's see how this affects the quadratic equation. Since c > 0, the quadratic at x=0 is positive. But at x=1, the quadratic is a + b + c < 0, so f(1) < 0. Therefore, the quadratic crosses the x-axis between 0 and 1. So there's at least one real root there, so discriminant is at least zero. But we need to show it's strictly greater. If the quadratic only touches the x-axis (discriminant zero), then it would have a repeated root. But if there's a sign change between 0 and 1, and the quadratic is continuous, then even if there is a repeated root, it would have to be at some point where the function doesn't cross the axis, but just touches it. But wait, if there's a sign change, then it must cross the axis, meaning there must be two distinct roots. Wait, but a quadratic can have a repeated root and still cross the axis. Wait, no. If a quadratic has a repeated root, it touches the axis at that root but doesn't cross it. So if there is a sign change, that would imply that the quadratic actually crosses the axis, hence two distinct roots, hence discriminant positive. Therefore, in Case 1, discriminant must be positive.Similarly, in Case 2: c < 0, so f(0) = c < 0. And f(1) = a + b + c > 0. So the quadratic goes from negative at x=0 to positive at x=1, hence crosses the x-axis between 0 and 1. Again, this would imply two distinct real roots (since a quadratic can't have more than two), so discriminant must be positive.Therefore, in both cases, the discriminant is positive. Hence, proved.Wait, but maybe I need to formalize this argument. Let me think again.Given that f(0) = c and f(1) = a + b + c. The product f(0)*f(1) < 0, so f changes sign between 0 and 1. By the Intermediate Value Theorem, there exists at least one root in (0,1). Since quadratics are continuous and differentiable, and can have at most two real roots. If there is one root in (0,1), then where is the other root? Well, if the quadratic crosses the x-axis once, it must cross again unless it's a tangent, but the discriminant being zero would mean a repeated root. However, if the quadratic has a repeated root, then it doesn't cross the axis but touches it. But if there's a sign change, then it must cross the axis, so there must be two distinct real roots, hence discriminant positive.Alternatively, suppose the discriminant is zero. Then the quadratic has a repeated root r. So f(x) = a(x - r)^2. Then f(0) = a*r², and f(1) = a(1 - r)^2. Since a is not zero (otherwise, the quadratic becomes linear). Then the product f(0)*f(1) = a² r² (1 - r)^2, which is non-negative. But the given condition is that f(0)*f(1) < 0, which is a contradiction. Therefore, discriminant cannot be zero. Hence discriminant must be positive.Therefore, regardless of the case, discriminant is positive. Hence, proved.Another approach could be algebraic. Let's start with (a + b + c)c < 0. Let's expand that:ac + bc + c² < 0.Rearranged: ac + bc + c² < 0.Factor out c: c(a + b + c) < 0, which is the original condition.But how does this relate to the discriminant? The discriminant is b² - 4ac. Let's see if we can relate these terms.From the inequality c(a + b + c) < 0, we have that c(a + b + c) is negative. Let's denote D = b² - 4ac. We need to show D > 0.Let me try to express D in terms of the given inequality. Since D = b² - 4ac, maybe we can relate this to the given inequality.Let me solve the given inequality for ac:From c(a + b + c) < 0,ac + bc + c² < 0,So ac < -bc - c².Then, substituting into D:D = b² - 4ac > b² - 4(-bc - c²) = b² + 4bc + 4c² = (b + 2c)².So D > (b + 2c)².But since (b + 2c)² is always non-negative, this shows that D is greater than or equal to zero. But we need D > 0. Wait, but (b + 2c)² is non-negative, so D > non-negative. Hmm, but this doesn't necessarily ensure D is positive. Because if (b + 2c)² is zero, then D > 0 would still hold? Wait, if (b + 2c)² = 0, then b + 2c = 0. Then D > 0 would imply that 0 > 0, which is a contradiction. Wait, so perhaps this approach is flawed.Alternatively, maybe we can use the Cauchy-Schwarz inequality or another inequality. Let me think.Alternatively, let's consider that since c(a + b + c) < 0, then either c > 0 and a + b + c < 0, or c < 0 and a + b + c > 0. Let's take the first case: c > 0 and a + b + c < 0. Then, since a + b + c < 0, we can write a + b < -c. Then, let's see if we can find a relation for the discriminant.Alternatively, consider the quadratic f(x) = ax² + bx + c. We know that f(0) = c and f(1) = a + b + c. The product f(0)f(1) < 0 implies that there is a root between 0 and 1. Similarly, if we look at the behavior of the quadratic as x approaches infinity, depending on the sign of a, the quadratic tends to positive or negative infinity. Let's consider that.Suppose a > 0. Then as x approaches infinity, f(x) tends to infinity. If there is a root between 0 and 1, then depending on the sign of f(0), which is c, we can see if there's another root.Wait, if a > 0 and c > 0 (Case 1), but f(1) = a + b + c < 0. So starting from x=0, f(0) = c > 0, then goes down to f(1) < 0, crosses the x-axis between 0 and 1, then since as x approaches infinity, f(x) tends to infinity, it must come back up and cross the x-axis again somewhere beyond x=1. Therefore, two real roots, hence discriminant positive.Similarly, if a > 0 and c < 0 (Case 2), then f(0) = c < 0 and f(1) = a + b + c > 0. So it crosses from negative to positive between 0 and 1, and as x approaches negative infinity, f(x) tends to positive infinity (since a > 0 and x² dominates). Therefore, it might cross the x-axis again somewhere before x=0. Hence two real roots, discriminant positive.If a < 0, similar reasoning applies. If a < 0 and c > 0 (Case 1), f(0) = c > 0, f(1) < 0. Since a < 0, as x approaches infinity, f(x) tends to negative infinity. So the graph goes from positive at x=0, negative at x=1, then continues to negative infinity. So it might cross the x-axis once between 0 and 1, but since a < 0, the parabola opens downward, so if there's a root between 0 and 1, then maybe another root to the left of x=0. Let's check: as x approaches negative infinity, f(x) tends to negative infinity (since a < 0). So from negative infinity, comes up to f(0) = c > 0, crosses the x-axis somewhere left of x=0, then goes up to x=0 at c > 0, then down to f(1) < 0, crossing again between 0 and 1. Wait, but if a < 0, the parabola opens downward. So if f(0) = c > 0 and f(1) < 0, then the maximum of the parabola is somewhere between x=0 and x=1. Wait, actually, the vertex of the parabola is at x = -b/(2a). Depending on where the vertex is, the quadratic could cross the x-axis twice. Wait, but regardless of the direction, if there's a sign change between f(0) and f(1), and the function tends to opposite infinities at the extremes, there must be two real roots. Hence discriminant positive.Similarly, in Case 2 where c < 0 and a + b + c > 0 with a < 0: f(0) = c < 0, f(1) > 0. As x approaches infinity, f(x) tends to negative infinity. So the function starts at x=0 with negative value, goes up to positive at x=1, then back down to negative infinity. So crosses the x-axis between 0 and 1, and then again somewhere beyond x=1. Hence two real roots, discriminant positive.Therefore, regardless of the value of a, the discriminant must be positive.Alternatively, another approach is to consider the quadratic equation and use the fact that if f(0) and f(1) have opposite signs, then there's a root between 0 and 1, and since quadratics can have at most two roots, the existence of one root implies another, hence discriminant positive. But I need to make sure that a quadratic with one root (discriminant zero) cannot have f(0) and f(1) with opposite signs. But as earlier, if discriminant is zero, then f(x) = a(x - r)^2. Then f(0) = a r² and f(1) = a (1 - r)^2. Both f(0) and f(1) would have the same sign as a, since squares are non-negative. Hence, their product would be non-negative. But the problem states that their product is negative. Therefore, discriminant cannot be zero, so discriminant must be positive.This seems to be a solid argument. Let me summarize:1. Given (a + b + c)c < 0, which implies f(1)f(0) < 0 where f(x) = ax² + bx + c.2. By the Intermediate Value Theorem, there's a root between 0 and 1.3. If discriminant were zero, f(x) would have a repeated root and f(0) and f(1) would both be non-negative or non-positive, contradicting f(0)f(1) < 0.4. Therefore, discriminant cannot be zero, so it must be positive.Hence, proved that b² - 4ac > 0.Alternatively, another algebraic approach: Suppose that (a + b + c)c < 0. Let me consider this inequality:(a + b + c)c < 0Let me denote S = a + b + c. Then Sc < 0. So either S > 0 and c < 0, or S < 0 and c > 0.Let me consider both cases.Case 1: S > 0 and c < 0.Since S = a + b + c > 0 and c < 0, then a + b = S - c. Since c is negative, subtracting a negative is adding a positive. So a + b = S + |c|. But not sure if helpful.Alternatively, since c < 0, and S > 0, then a + b + c > 0, but c < 0. Maybe express a as a = S - b - c. Then substitute into discriminant:D = b² - 4ac = b² - 4(S - b - c)c = b² - 4Sc + 4bc + 4c².But since S > 0 and c < 0, -4Sc is positive (since negative times negative). 4bc could be positive or negative, and 4c² is positive. So D is a sum of positive terms and possibly some others. Not sure.Alternatively, maybe use Cauchy-Schwarz or another inequality.Alternatively, consider that since f(0)f(1) < 0, then there's a root between 0 and 1. Let me call that root α. Then, the quadratic can be written as f(x) = a(x - α)(x - β), where β is the other root.Then, expanding this, f(x) = a x² - a(α + β)x + a α β.Comparing coefficients:b = -a(α + β)c = a α βTherefore, the discriminant D = b² - 4ac = [ -a(α + β) ]² - 4a * a α β = a² (α + β)² - 4a² α β = a² [ (α + β)^2 - 4αβ ] = a² (α - β)^2.Since a is real and non-zero (if a were zero, the equation is linear, but then (a + b + c)c becomes (b + c)c, and the problem states a, b, c are real numbers, but doesn't specify a ≠ 0. Wait, hold on, if a = 0, then the equation is linear, bx + c = 0. Then the discriminant is not b² - 4ac but rather just b², but the problem statement says "for real numbers a, b, c", so maybe a could be zero. Wait, but if a is zero, the discriminant becomes b² - 0 = b². If a = 0, then (a + b + c)c = (b + c)c < 0. So either (b + c) and c have opposite signs. If (b + c)c < 0, then c ≠ 0. Then, for linear equation bx + c = 0, the root is x = -c/b. But in this case, if a = 0, then discriminant D = b². If a = 0 and D = b² > 0, unless b = 0. But if a = 0 and b = 0, then the equation becomes c = 0, but then (a + b + c)c = 0 * c = 0, which contradicts the given (a + b + c)c < 0. So if a = 0, then b can't be zero, so discriminant is b² > 0. Therefore, even if a = 0, the discriminant is positive. Wait, but the problem didn't specify that a ≠ 0. So maybe I need to consider the case when a = 0 separately.But in the original problem statement, it just says "for real numbers a, b, c", not necessarily that a ≠ 0. So if a = 0, then the quadratic becomes linear. However, in that case, the discriminant of the quadratic equation (which would be b² - 4ac = b²) is still positive unless b = 0. But if a = 0 and b = 0, then the equation is c = 0, which would make (a + b + c)c = 0 * 0 = 0, contradicting the given inequality. Therefore, if a = 0, then b ≠ 0, so discriminant is b² > 0. Hence, in all cases, discriminant is positive.But wait, the problem states "for real numbers a, b, c", but in the case when a = 0, it's a linear equation. The discriminant terminology is typically used for quadratics, but the problem statement still refers to b² - 4ac. So even if a = 0, we can still compute b² - 4ac, which is b². So in that case, the discriminant being positive is equivalent to b ≠ 0. But given that (a + b + c)c < 0, if a = 0, then (0 + b + c)c < 0 ⇒ (b + c)c < 0. So (b + c)c < 0 implies that c ≠ 0 and b + c and c have opposite signs. So if c > 0, then b + c < 0 ⇒ b < -c. If c < 0, then b + c > 0 ⇒ b > -c. In either case, since c ≠ 0, and b is related to c, but does this imply that b ≠ 0? Let's see.Suppose a = 0, c > 0, and b + c < 0. Then b < -c. Since c > 0, -c < 0, so b < -c < 0. Therefore, b is negative, so b ≠ 0. Similarly, if a = 0, c < 0, then b + c > 0 ⇒ b > -c. Since c < 0, -c > 0, so b > 0. Therefore, b ≠ 0. Hence, in the case a = 0, discriminant D = b² > 0. So even when a = 0, the discriminant is positive. Therefore, regardless of whether a is zero or not, the discriminant is positive.Therefore, the original statement holds even if a = 0. But in most discussions about quadratic equations, a is non-zero, but since the problem allows a to be any real number, we have to consider that case.However, in the problem statement, the discriminant is given as b² - 4ac, which is the standard discriminant for a quadratic equation. But if a = 0, then the equation is linear, and the discriminant as defined here is still b². Therefore, even in that case, the discriminant is positive because (a + b + c)c < 0 forces b ≠ 0 as shown above.Therefore, whether a is zero or not, the discriminant is positive. Therefore, the proof holds.Another way to think about it is using the properties of quadratic functions. If a quadratic function f(x) = ax² + bx + c has f(0) and f(1) with opposite signs, then it must cross the x-axis at least once between 0 and 1. If a ≠ 0, then being a quadratic, it can have at most two real roots. But since it's crossing once between 0 and 1, and depending on the leading coefficient a, the function will tend to ±infinity as x approaches ±infinity. Therefore, there must be another real root, making the total number of real roots two, hence discriminant positive.If a = 0, as discussed, the equation is linear with a non-zero slope (since b ≠ 0), so it has exactly one real root. However, the discriminant in this case is b², which is positive because b ≠ 0. Therefore, in all cases, discriminant is positive.Thus, combining all these observations, the discriminant b² - 4ac must indeed be greater than zero under the given condition.

✅:To prove that ( b^2 - 4ac > 0 ) given ( (a + b + c)c < 0 ) for real numbers ( a, b, c ), we analyze the quadratic function ( f(x) = ax^2 + bx + c ):1. Interpret the Given Inequality: - The condition ( (a + b + c)c < 0 ) implies ( f(1) cdot f(0) < 0 ), where ( f(1) = a + b + c ) and ( f(0) = c ). - This means ( f(1) ) and ( f(0) ) have opposite signs.2. Intermediate Value Theorem: - Since ( f(x) ) is continuous and changes sign between ( x = 0 ) and ( x = 1 ), there exists at least one real root in ( (0, 1) ).3. Discriminant Analysis: - For ( f(x) ) to have real roots, the discriminant ( D = b^2 - 4ac ) must be non-negative. However, we need to show ( D > 0 ). - If ( D = 0 ), ( f(x) ) would have a repeated root ( r ), and ( f(x) = a(x - r)^2 ). Evaluating at ( x = 0 ) and ( x = 1 ): [ f(0) = ar^2 quad text{and} quad f(1) = a(1 - r)^2. ] Both ( f(0) ) and ( f(1) ) would share the sign of ( a ), contradicting ( f(0)f(1) < 0 ). Thus, ( D neq 0 ).4. Case Analysis: - Case 1: ( c > 0 ). Then ( f(0) > 0 ) and ( f(1) = a + b + c < 0 ). The quadratic must cross the x-axis twice (once in ( (0, 1) )), so ( D > 0 ). - Case 2: ( c < 0 ). Then ( f(0) < 0 ) and ( f(1) = a + b + c > 0 ). Similarly, the quadratic crosses the x-axis twice, ensuring ( D > 0 ).5. Handling ( a = 0 ): - If ( a = 0 ), the equation becomes linear ( bx + c = 0 ). The discriminant simplifies to ( b^2 ), which is positive since ( (a + b + c)c < 0 ) forces ( b neq 0 ).Conclusion: In all scenarios, ( b^2 - 4ac > 0 ) holds. Thus, the discriminant is necessarily positive under the given condition.[boxed{b^2 - 4ac > 0}]

🔑:1. Consider the quadratic trinomial ( f(x) = x^2 + bx + ac ). We want to analyze the properties of this quadratic equation. 2. Evaluate ( f(x) ) at ( x = c ): [ f(c) = c^2 + bc + ac ] By factoring, we notice: [ f(c) = c^2 + c(b + a) ] Given our initial condition: [ (a + b + c)c < 0 ] Thus: [ f(c) = (a + b + c)c < 0 ] This tells us ( f(c) ) is negative.3. Given that the coefficient of ( x^2 ) in ( f(x) ) is positive, i.e., ( 1 > 0 ): Since the coefficient of ( x^2 ) is positive, the parabola opens upwards. For ( f(c) ) to be negative, it must cross the ( x )-axis, meaning the quadratic equation has real roots. 4. Determine the condition for a quadratic to have real roots: A necessary and sufficient condition for a quadratic equation ( ax^2 + bx + c ) to have real roots is that its discriminant ( Delta ) be non-negative, i.e., [ Delta = b^2 - 4ac geq 0 ]5. However, since ( f(c) < 0 ) and the parabola must cross the ( x )-axis, the discriminant must actually be positive: [ Delta = b^2 - 4ac > 0 ]6. Therefore, we have shown that the discriminant of the quadratic ( f(x) ) is positive. Conclusion:[boxed{b^2 - 4ac > 0}]

🧠:Okay, let's try to tackle this problem. So, we have two distinct numbers x and y that satisfy the equation x² - 2000x = y² - 2000y. We need to find the sum of x and y. Hmm, let's see. First, I remember that when you have equations involving squares of variables, sometimes rearranging terms or factoring can help. Let me try to rearrange the equation first. The equation is x² - 2000x = y² - 2000y. Maybe I can bring all the terms to one side to see if something cancels out or factors nicely. Let's subtract y² and add 2000y to both sides:x² - 2000x - y² + 2000y = 0.Hmm, that gives x² - y² - 2000x + 2000y = 0. Now, I notice that x² - y² is a difference of squares, which factors into (x - y)(x + y). Let's try that:(x - y)(x + y) - 2000(x - y) = 0.Wait, because -2000x + 2000y is the same as -2000(x - y). So, factoring out (x - y) from both terms:(x - y)[(x + y) - 2000] = 0.Okay, so this product equals zero, which means either (x - y) = 0 or (x + y - 2000) = 0. But the problem states that x and y are distinct numbers, so x - y cannot be zero. Therefore, the other factor must be zero:x + y - 2000 = 0 => x + y = 2000.Wait, is that all? That seems straightforward, but let me double-check to make sure I didn't miss anything. Let me retrace my steps:Original equation: x² - 2000x = y² - 2000y.Rearranged: x² - y² = 2000x - 2000y.Factored left side: (x - y)(x + y) = 2000(x - y).Then, since x ≠ y, we can divide both sides by (x - y), resulting in x + y = 2000.Yes, that seems correct. So, the sum of x and y is 2000. But wait, let me test with some numbers to see if this holds. Suppose x + y = 2000, then let's pick x = 1000 and y = 1000. But they are not distinct. So, let's pick x = 1001 and y = 999. Then x + y = 2000. Let's check the original equation:Left side: 1001² - 2000*1001 = (1001)^2 - 2000*1001.Right side: 999² - 2000*999.Calculate both sides:Left: 1001*(1001 - 2000) = 1001*(-999) = -999999.Right: 999*(999 - 2000) = 999*(-1001) = -999999.Yes, they are equal. So even though x and y are distinct, their sum is 2000. That works. So, the answer is 2000. Hmm, seems straightforward once you factor it. But let me think if there's another approach in case I missed something.Alternatively, maybe completing the square. Let's see. Let's take the original equation:x² - 2000x = y² - 2000y.Move all terms to the left:x² - y² - 2000x + 2000y = 0.Wait, that's the same as before. Alternatively, complete the square for both x and y.Let me try that. For x: x² - 2000x. To complete the square, take half of 2000, which is 1000, square it to get 1000000. So:x² - 2000x = (x - 1000)^2 - 1000000.Similarly for y: y² - 2000y = (y - 1000)^2 - 1000000.So substituting back into the original equation:(x - 1000)^2 - 1000000 = (y - 1000)^2 - 1000000.Cancel out the -1000000 on both sides:(x - 1000)^2 = (y - 1000)^2.Taking square roots on both sides, we get:x - 1000 = ±(y - 1000).So, two cases:1. x - 1000 = y - 1000 => x = y. But x and y are distinct, so discard this.2. x - 1000 = -(y - 1000) => x - 1000 = -y + 1000 => x + y = 2000.Exactly the same result. So this confirms that x + y = 2000. Therefore, regardless of the method, the sum is 2000. Wait, but maybe there's a possibility that even when x ≠ y, there's another solution? Let me think. Suppose x + y ≠ 2000, but then the original equation would not hold. Let's suppose x + y = k. Then from the first method, we had (x - y)(k - 2000) = 0. Since x ≠ y, then k must be 2000. So no other solutions. Therefore, the sum must indeed be 2000.I think that's solid. So the answer is 2000. The sum of x and y is 2000.

✅:To solve the equation ( x^2 - 2000x = y^2 - 2000y ) for distinct numbers ( x ) and ( y ), we start by rearranging and factoring:1. Subtract ( y^2 ) and add ( 2000y ) to both sides: [ x^2 - y^2 - 2000x + 2000y = 0. ]2. Factor the difference of squares and group terms: [ (x - y)(x + y) - 2000(x - y) = 0. ]3. Factor out ( (x - y) ): [ (x - y)(x + y - 2000) = 0. ]Since ( x neq y ), the factor ( (x - y) neq 0 ). Thus, the remaining factor must be zero: [ x + y - 2000 = 0 implies x + y = 2000. ]Verification via completing the square confirms the result. Let ( x + y = 2000 ), and distinct values like ( x = 1001 ), ( y = 999 ) satisfy the original equation. Hence, the sum of ( x ) and ( y ) is:[boxed{2000}]

🔑:1. Start with the given equation: [ x^2 - 2000x = y^2 - 2000y ]2. Rearrange the terms to isolate the quadratic and linear parts: [ x^2 - y^2 = 2000x - 2000y ]3. Factor the left-hand side using the difference of squares formula and factor out common terms on the right-hand side: [ (x - y)(x + y) = 2000(x - y) ]4. Given that ( x neq y ), the ((x - y)) term on both sides can be canceled out: [ x + y = 2000 ]5. Hence, the sum of the numbers (x) and (y) is: [ boxed{2000} ]

🧠:Okay, so I need to prove that if m and n are natural numbers with m less than sqrt(2) times n, then m/n is less than sqrt(2) multiplied by (1 minus 1/(4n²)). Hmm, let me start by understanding what the problem is asking. First, we have m and n as positive integers, and m is less than sqrt(2) times n. The goal is to show that the fraction m/n is bounded above not just by sqrt(2), but actually by a slightly smaller value, which is sqrt(2) times (1 - 1/(4n²)). So this upper bound depends on n, getting closer to sqrt(2) as n increases because the term 1/(4n²) becomes smaller. I need to figure out why subtracting that 1/(4n²) term is necessary. Maybe it's because m/n can't get too close to sqrt(2) when m and n are integers. Since m is an integer less than sqrt(2)n, perhaps the difference between sqrt(2)n and m is at least some minimal amount, which when translated into the fraction m/n gives us this bound. Let me try to approach this step by step. Let's denote that m < sqrt(2) n. Since m and n are natural numbers, m is the greatest integer less than sqrt(2) n, or possibly something else? Wait, no, m can be any natural number less than sqrt(2) n. So the maximum possible value of m is the floor of sqrt(2) n. So m ≤ floor(sqrt(2) n). Therefore, the maximum value of m/n is floor(sqrt(2) n)/n. So we need to show that floor(sqrt(2) n)/n < sqrt(2)(1 - 1/(4n²)). Alternatively, maybe instead of considering the floor function, we can think about the difference between sqrt(2) n and m. Let me define k = sqrt(2) n - m. Since m is an integer less than sqrt(2) n, k must be a positive real number. Moreover, since m is an integer, the smallest possible value of k is when m is the largest integer less than sqrt(2) n, i.e., m = floor(sqrt(2) n). Then k would be sqrt(2) n - floor(sqrt(2) n), which is the fractional part of sqrt(2) n. But how does this help? The fractional part of sqrt(2) n is between 0 and 1. But I need to relate k to the bound given. Let's see. If k = sqrt(2) n - m, then m/n = sqrt(2) - k/n. So we need to show that sqrt(2) - k/n < sqrt(2)(1 - 1/(4n²)). Rearranging this inequality: sqrt(2) - k/n < sqrt(2) - sqrt(2)/(4n²). Subtracting sqrt(2) from both sides gives -k/n < -sqrt(2)/(4n²). Multiplying both sides by -1 (which reverses the inequality) gives k/n > sqrt(2)/(4n²). Then multiplying both sides by n gives k > sqrt(2)/(4n). So the problem reduces to showing that k, which is sqrt(2) n - m, is greater than sqrt(2)/(4n). Therefore, if I can show that the difference between sqrt(2) n and the next integer m is at least sqrt(2)/(4n), then the original inequality will hold. But how can I prove that k > sqrt(2)/(4n)? Since sqrt(2) is irrational, the fractional parts of sqrt(2) n are distributed in a certain way, but maybe we can use some properties of Diophantine approximation here. Specifically, maybe we can use the fact that sqrt(2) can't be approximated too well by rationals. The classic result is that for any irrational number α, there are infinitely many fractions p/q such that |α - p/q| < 1/(sqrt(5) q²), but for sqrt(2), the approximation constant is known to be 1/(2sqrt(2)). Wait, perhaps even better. Let me recall that for quadratic irrationals, the continued fraction expansion is periodic, so sqrt(2) has a continued fraction [1; (2)], repeating 2 indefinitely. The convergents of the continued fraction give the best approximations. But perhaps instead of going into continued fractions, which might be more complex, I can approach this using the idea that if m/n is close to sqrt(2), then |sqrt(2) - m/n| must be at least some lower bound. In general, for any rational approximation m/n to sqrt(2), we have |sqrt(2) - m/n| ≥ 1/(n^2 (sqrt(2) + 2)). Wait, is that a standard bound? Let me think. By the theory of continued fractions, for a continued fraction convergent p_k/q_k of α, we have |α - p_k/q_k| < 1/(q_k q_{k+1}). For sqrt(2), the convergents have q_{k+1} = 2 q_k + q_{k-1}, but perhaps it's more straightforward to use a general result. Alternatively, suppose that m/n is a rational approximation to sqrt(2). Then cross-multiplying, we can consider |sqrt(2) n - m|. Let me denote this as k, as before. So k = |sqrt(2) n - m|. Since m and n are integers, k is the absolute difference between sqrt(2) n and the nearest integer. For irrational numbers like sqrt(2), Dirichlet's approximation theorem tells us that there are infinitely many n such that k < 1/n. But we need a lower bound here, not an upper bound. Wait, but in our problem, we need to show that k > sqrt(2)/(4n). So for all n, the difference between sqrt(2) n and the nearest integer m is at least sqrt(2)/(4n). But is that true? Let's test with small n. Take n = 1: sqrt(2) ≈ 1.414. The nearest integer m is 1. So k ≈ 0.414. sqrt(2)/(4*1) ≈ 0.353. 0.414 > 0.353, so holds. n = 2: sqrt(2)*2 ≈ 2.828. Nearest integer m is 3, but wait, m must be less than sqrt(2)*n. So m must be less than 2.828, so m=2. Then k ≈ 0.828. sqrt(2)/(4*2) ≈ 0.353/2 ≈ 0.176. 0.828 > 0.176, holds. n = 3: sqrt(2)*3 ≈ 4.242. m must be less than 4.242, so m=4. Then k ≈ 0.242. sqrt(2)/(4*3) ≈ 0.353/3 ≈ 0.117. 0.242 > 0.117, holds. n = 5: sqrt(2)*5 ≈ 7.071. m=7, k≈0.071. sqrt(2)/(4*5) ≈ 0.353/5 ≈ 0.0706. So here, k ≈0.071 which is just above 0.0706. So it holds. n=7: sqrt(2)*7≈9.899. m=9, k≈0.899. sqrt(2)/(4*7)=approx 0.353/7≈0.0504. 0.899>0.0504, holds. n=12: sqrt(2)*12≈16.97. m=16, k≈0.97. sqrt(2)/(4*12)=0.353/12≈0.0294. 0.97>0.0294, holds. Wait, n=5 was the closest. Let's check n=5 more precisely. sqrt(2)*5 ≈7.0710678. So m=7, which is less than 7.071..., so k=7.0710678 -7≈0.0710678. sqrt(2)/(4*5)=sqrt(2)/20≈0.0707107. So 0.0710678 > 0.0707107. Indeed, barely, but it's true. So in that case, for n=5, k is approximately 0.0710678, which is just barely larger than sqrt(2)/20 ≈0.0707107. So this suggests that the bound k > sqrt(2)/(4n) holds for all natural numbers n, which is exactly what we need. Therefore, if we can establish that for any natural number n, the difference k = sqrt(2) n - m (where m is the largest integer less than sqrt(2) n) is greater than sqrt(2)/(4n), then we can substitute that into our inequality and get the desired result. But how to prove that k > sqrt(2)/(4n)? Let's suppose for contradiction that k ≤ sqrt(2)/(4n). Then we have:sqrt(2) n - m ≤ sqrt(2)/(4n)Multiply both sides by n:sqrt(2) n² - m n ≤ sqrt(2)/4Rearranged:sqrt(2) n² - sqrt(2)/4 ≤ m nBut m is an integer less than sqrt(2) n. So m n is less than sqrt(2) n². Therefore:sqrt(2) n² - sqrt(2)/4 ≤ m n < sqrt(2) n²Which implies:sqrt(2) n² - sqrt(2)/4 < sqrt(2) n²Which is true, but not helpful. Maybe another approach.Alternatively, consider the inequality |sqrt(2) - m/n| < sqrt(2)/(4n²). If we can show that this inequality cannot hold for any integers m, n, then the reverse inequality must hold, which would give us |sqrt(2) - m/n| ≥ sqrt(2)/(4n²), which would lead to the desired result. So suppose that |sqrt(2) - m/n| < sqrt(2)/(4n²). Let's square both sides to eliminate the square root. However, squaring might complicate things, but let's try.Left side squared: (sqrt(2) - m/n)^2 < (sqrt(2)/(4n²))^2Expand the left side:2 - 2 sqrt(2) (m/n) + (m/n)^2 < 2/(16n^4)Simplify:2 - 2 sqrt(2) (m/n) + (m²)/(n²) < 1/(8n^4)Multiply both sides by n² to eliminate denominators:2n² - 2 sqrt(2) m n + m² < 1/(8n²)But the left side is 2n² - 2 sqrt(2) m n + m², which is a quadratic in m. However, since m is an integer, maybe we can find a contradiction here. Alternatively, cross-multiplying the original inequality:|sqrt(2) - m/n| < sqrt(2)/(4n²)Multiply both sides by n:|sqrt(2) n - m| < sqrt(2)/(4n)But |sqrt(2) n - m| is k, which is at least 1/(some expression). Wait, but sqrt(2) is irrational, so sqrt(2) n - m can't be zero. The minimal possible value of |sqrt(2) n - m| is related to how well sqrt(2) can be approximated by rationals. Perhaps a better approach is to use the mean value theorem or consider the function f(n) = sqrt(2) n - m and relate it to derivatives? Wait, maybe not. Alternatively, consider that sqrt(2) is irrational, so there's no exact solution, but I need a lower bound on the difference. Alternatively, suppose that m/n is a rational approximation to sqrt(2). Then, cross-multiplying:m ≈ sqrt(2) nSquaring both sides:m² ≈ 2 n²But since m and n are integers, m² - 2n² must be a non-zero integer. The minimal absolute value of m² - 2n² is 1, known from Pell's equation. Wait, Pell's equation is x² - 2 y² = ±1. So for solutions to x² - 2 y² = ±1, which are the convergents of the continued fraction expansion of sqrt(2). Indeed, the fundamental solution is (3, 2) since 3² - 2*2² = 9 - 8 = 1. Then solutions can be generated by (3 + 2 sqrt(2))^k. But if m² - 2n² = ±1, then |m² - 2n²| = 1. Let's see how that helps. From the equation m² - 2n² = ±1, we can write:(m/n)² = 2 ± 1/n²Taking square roots:m/n = sqrt(2 ± 1/n²)But expanding sqrt(2 ± 1/n²) using Taylor series, perhaps. Let's approximate:sqrt(2 + ε) ≈ sqrt(2) + ε/(2 sqrt(2)) - ε²/(8 * (2)^(3/2)) + ...Similarly, sqrt(2 - ε) ≈ sqrt(2) - ε/(2 sqrt(2)) - ε²/(8 * (2)^(3/2)) + ...So for ε = 1/n², we have:If m² - 2n² = 1, then m/n = sqrt(2 + 1/n²) ≈ sqrt(2) + 1/(2 sqrt(2) n²) - ... Similarly, if m² - 2n² = -1, then m/n = sqrt(2 - 1/n²) ≈ sqrt(2) - 1/(2 sqrt(2) n²) - ... But in our problem, m/n is less than sqrt(2). So the case m² - 2n² = -1 would give m/n = sqrt(2 - 1/n²), which is less than sqrt(2). But perhaps this is the closest approximation from below. So for the convergents of sqrt(2), which are the solutions to Pell's equation, the fractions m/n approximate sqrt(2) alternately from above and below. But in our case, since m < sqrt(2) n, we're looking at approximations from below. So if m/n is the closest fraction to sqrt(2) below sqrt(2), then m/n = sqrt(2) - k/n, where k is as before. But if m/n is a convergent of sqrt(2), then we have |sqrt(2) - m/n| < 1/(2n²). Wait, is that true? Let me check. The convergents p/q of a continued fraction satisfy |α - p/q| < 1/(q² a_{k+1}), where a_{k+1} is the next term in the continued fraction. For sqrt(2), the continued fraction is [1; 2,2,2,...], so all a_k after the first are 2. Therefore, |sqrt(2) - p/q| < 1/(q² * 2). Therefore, for convergents, the error is less than 1/(2 q²). But in our problem, we need a lower bound. So, for non-convergents, perhaps the error is larger. But even so, if a convergent satisfies |sqrt(2) - p/q| < 1/(2 q²), then other fractions would have larger errors. But in our problem, we need to show that for any m/n < sqrt(2), sqrt(2) - m/n > sqrt(2)/(4 n²). Wait, rearranged, this is:sqrt(2) - m/n > sqrt(2)/(4 n²)Multiply both sides by n:sqrt(2) n - m > sqrt(2)/(4 n)But we need to show that the difference sqrt(2) n - m is greater than sqrt(2)/(4 n). If m/n is a convergent of sqrt(2), then |sqrt(2) - m/n| < 1/(2 n²). So in that case:sqrt(2) - m/n < 1/(2 n²)But this is an upper bound, not a lower bound. However, if it's not a convergent, then the error is larger. But in our problem, we need a lower bound for all m/n < sqrt(2). So even if m/n is a convergent, which has a very small error, we need to show that sqrt(2) - m/n > sqrt(2)/(4 n²). But wait, for convergents, we have sqrt(2) - m/n < 1/(2 n²), but we need to compare this to sqrt(2)/(4 n²). Since 1/(2 n²) ≈ 0.5/n² and sqrt(2)/(4 n²) ≈ 0.353/n². So 1/(2 n²) is larger than sqrt(2)/(4 n²). Therefore, if the convergent satisfies sqrt(2) - m/n < 1/(2 n²), which is bigger than sqrt(2)/(4 n²), that doesn't directly help. Wait, but if we have sqrt(2) - m/n < 1/(2 n²), then it's possible that sqrt(2) - m/n is less than sqrt(2)/(4 n²). But we need to show that sqrt(2) - m/n > sqrt(2)/(4 n²). That would not hold if there exists a convergent where sqrt(2) - m/n < sqrt(2)/(4 n²). But wait, 1/(2 n²) vs sqrt(2)/(4 n²). Since sqrt(2)/4 ≈ 0.353, and 1/2 = 0.5. So 1/(2 n²) > sqrt(2)/(4 n²). Therefore, if sqrt(2) - m/n < 1/(2 n²), it could still be greater than sqrt(2)/(4 n²). For example, take n=5. The convergent is m/n=7/5=1.4. sqrt(2)-7/5≈1.4142-1.4≈0.0142. sqrt(2)/(4*25)=1.4142/(100)=0.014142. So in this case, sqrt(2) - m/n ≈0.0142 >0.014142. So it's slightly larger. Similarly, for n=12. The convergent is 17/12≈1.4167. sqrt(2)≈1.4142, so sqrt(2) -17/12≈-0.0025. Wait, that's negative. Wait, but 17/12 is greater than sqrt(2). So actually, the convergents alternate between being above and below sqrt(2). So for even k, convergents are below sqrt(2), and for odd k, they're above. Wait, let's check. The continued fraction for sqrt(2) is [1; 2,2,2,...]. The convergents are:a0 = 1/1 = 1a1 = 3/2 = 1.5a2 = 7/5 = 1.4a3 = 17/12 ≈1.4167a4 = 41/29 ≈1.4138a5 = 99/70 ≈1.4142857So alternates above and below. So the even-indexed convergents (starting from 0) are below, odd are above. So for example, a0=1 < sqrt(2), a1=3/2 > sqrt(2), a2=7/5 < sqrt(2), a3=17/12 > sqrt(2), etc. Therefore, for the convergents below sqrt(2), like a0, a2, a4, etc., we can check the difference. For a2=7/5: sqrt(2)-7/5≈1.4142-1.4≈0.0142. sqrt(2)/(4*25)=≈0.014142. So indeed, 0.0142 >0.014142.For a4=41/29≈1.4138. sqrt(2)-41/29≈1.4142 -1.4138≈0.0004. sqrt(2)/(4*29²)=sqrt(2)/(4*841)=≈1.4142/3364≈0.000420. So here, 0.0004 <0.000420. Wait, that contradicts our previous assertion. Wait, sqrt(2)/(4n²) for n=29 would be sqrt(2)/(4*841)=sqrt(2)/3364≈0.000420. But sqrt(2)-41/29≈0.0004, which is less than that. So in this case, the difference is smaller than sqrt(2)/(4n²). That would violate the inequality we're supposed to prove. But wait, the problem statement says m and n are natural numbers with m < sqrt(2) n. So for n=29, m=41, since 41 < sqrt(2)*29≈41.0122. So m=41, n=29. Then the difference is sqrt(2)*29 -41≈0.0122. Wait, but wait, I calculated sqrt(2)-41/29≈0.0004, but sqrt(2)*29 -41≈0.0122. Wait, there's confusion here. Wait, let me clarify:The difference sqrt(2)*n -m is k≈0.0122 for n=29, m=41.Then k/n≈0.0122/29≈0.00042. sqrt(2)/(4n²)=sqrt(2)/(4*841)=sqrt(2)/3364≈0.000420. So k/n≈0.00042, which is approximately equal to sqrt(2)/(4n²). So in this case, k/n is slightly larger than sqrt(2)/(4n²). Because sqrt(2)*29 -41≈0.0122, so k=0.0122, k/n≈0.0122/29≈0.0004206, and sqrt(2)/(4*29²)=≈0.000420. So 0.0004206 >0.000420. So just barely holds. So in this case, the inequality holds by a very small margin. Similarly, for n=169 (next convergent), m=239. sqrt(2)*169≈239.0001, so k≈0.0001. Then k/n≈0.0001/169≈0.00000059. sqrt(2)/(4n²)=sqrt(2)/(4*28561)=≈1.4142/114244≈0.0000124. Wait, 0.00000059 is less than 0.0000124. Wait, that would contradict the inequality. Wait, but in reality, sqrt(2)*169≈239.0001, so k=0.0001. Then k/n=0.0001/169≈5.917e-7. sqrt(2)/(4*169²)=sqrt(2)/(4*28561)=≈1.4142/114244≈1.238e-5. So 5.917e-7 <1.238e-5. That would mean the inequality fails. But that can't be, because the problem statement says it should hold for all natural numbers m,n with m < sqrt(2) n. But in this case, for n=169, m=239, which is less than sqrt(2)*169≈239.0001. Then according to the problem's inequality, m/n should be less than sqrt(2)*(1 -1/(4n²)). Let's compute both sides:m/n=239/169≈1.414201183431953.sqrt(2)*(1 -1/(4*169²))≈1.41421356237*(1 -1/(4*28561))≈1.41421356237*(1 -1/114244)≈1.41421356237*0.999991243≈1.41421356237 -1.41421356237*0.000008757≈1.41421356237 -1.238e-5≈1.41420118.So m/n≈1.414201183 and the upper bound is≈1.41420118. So the upper bound is very slightly less than m/n. Which would mean that the inequality m/n < sqrt(2)(1 -1/(4n²)) is violated. But that contradicts the problem statement. Therefore, either there is a mistake in my calculations or reasoning, or the problem statement might have additional constraints. Wait, the problem says m and n are natural numbers with m < sqrt(2) n. Prove that m/n < sqrt(2)(1 -1/(4n²)). But in this case, m=239 and n=169, m < sqrt(2) n (since sqrt(2)*169≈239.0001), so m=239 is indeed less than sqrt(2) n. Then according to the problem's inequality, 239/169 should be less than sqrt(2)*(1 -1/(4*169²)). But my calculation shows that 239/169≈1.414201183 and the right side≈1.41420118, so m/n is actually slightly larger than the right-hand side. This suggests that the inequality does not hold for n=169, m=239, which would contradict the problem statement. Therefore, my previous reasoning must have an error. Wait, let's recalculate the upper bound for n=169. sqrt(2) ≈1.41421356237Compute 1 -1/(4n²) where n=169:1/(4n²)=1/(4*28561)=1/114244≈8.756×10^-6Thus, 1 -8.756×10^-6≈0.9999912435Multiply by sqrt(2):sqrt(2)*0.9999912435≈1.41421356237*0.9999912435≈1.41421356237 -1.41421356237*0.0000087565≈1.41421356237 -1.238e-5≈1.41421356237 -0.00001238≈1.41420118Now, m/n=239/169≈1.414201183431953So m/n≈1.4142011834 versus upper bound≈1.41420118. Therefore, m/n is indeed slightly larger than the upper bound. This contradicts the inequality. Therefore, the problem statement might be incorrect, or I have a miscalculation. Wait, but 239/169 is exactly (sqrt(2) - ε), where ε is extremely small. In fact, 239/169 is a convergent of sqrt(2), and such convergents are known to alternate between being less than and greater than sqrt(2). Wait, but n=169 and m=239, which gives m/n≈1.414201183, which is less than sqrt(2)≈1.41421356237. So m/n is indeed less than sqrt(2). But when we compute the upper bound sqrt(2)(1 -1/(4n²)), we get≈1.41420118, which is very slightly less than m/n≈1.4142011834. Therefore, the inequality m/n < sqrt(2)(1 -1/(4n²)) is not true in this case. Therefore, either the problem statement is incorrect, or there's a restriction in the problem that I missed. Let me check the problem statement again. It says: Let m, n ∈ N and m < sqrt(2) n. Prove that m/n < sqrt(2)(1 -1/(4n²)). But according to the example with n=169 and m=239, this inequality is not true. Therefore, either my calculation is wrong, or the problem statement is false. Alternatively, maybe the problem has a different context. For instance, perhaps m and n are supposed to be positive integers with m/n in lowest terms, but even so, 239 and 169 are coprime (since 239 is a prime number and 239 does not divide 169), so the fraction is reduced. Alternatively, maybe the problem has a condition that n is not part of a solution to Pell's equation, but that seems complex. Alternatively, perhaps I made a miscalculation. Let me verify with higher precision. Let me compute m/n for n=169 and m=239:239 ÷ 169:169 ×1 =169, 239-169=70. Bring down 0: 700.169×4=676, 700-676=24. Bring down 0:240.169×1=169, 240-169=71. Bring down 0:710.169×4=676, 710-676=34. Bring down 0:340.169×2=338, 340-338=2. Bring down 0:20.169×0=0, 20 remains. Bring down 0:200.169×1=169, 200-169=31. Bring down 0:310.169×1=169, 310-169=141. Bring down 0:1410.169×8=1352, 1410-1352=58. Bring down 0:580.169×3=507, 580-507=73. Bring down 0:730.169×4=676, 730-676=54. Bring down 0:540.169×3=507, 540-507=33. Bring down 0:330.169×1=169, 330-169=161. Bring down 0:1610.This is getting tedious, but the decimal expansion of 239/169 is approximately 1.414201183431953...Now, let's compute sqrt(2)*(1 -1/(4*169²)) with more precision. First, compute 1/(4*169²):169²=28561, so 1/(4*28561)=1/114244≈0.0000087567626So 1 - 0.0000087567626≈0.9999912432374Now, multiply by sqrt(2). Let's compute sqrt(2) with high precision:sqrt(2)=1.4142135623730950488016887242097...Multiply by 0.9999912432374:First, compute 1.4142135623730950488016887242097 * 0.9999912432374This is equal to 1.4142135623730950488016887242097 - 1.4142135623730950488016887242097 * 0.0000087567626Compute the second term:1.4142135623730950488016887242097 * 0.0000087567626≈1.4142135623730950488016887242097 * 8.7567626×10^-6≈1.4142135623730950488016887242097 ×8.7567626×10^-6≈1.4142135623730950488016887242097 *0.0000087567626Let me compute this:1.4142135623730950488016887242097 ×0.0000087567626First, note that 1.414213562373095 ×0.0000087567626 ≈Compute 1.414213562373095 ×8.7567626 = then multiply by 10^-6.Compute 1.414213562373095 ×8 =11.313708498984761.414213562373095 ×0.7567626 ≈1.414213562373095 ×0.7=0.98994949366116651.414213562373095 ×0.0567626≈1.414213562373095 ×0.05=0.070710678118654751.414213562373095 ×0.0067626≈0.0095602414Adding up:0.9899494936611665 +0.07071067811865475=1.0606601717798212+0.0095602414≈1.0702204131798212Total:11.31370849898476 +1.0702204131798212≈12.383928912164581Multiply by10^-6:≈0.000012383928912164581Therefore, the second term≈0.000012383928912164581Subtract this from sqrt(2):1.4142135623730950488016887242097 -0.000012383928912164581≈1.4142011784441828848016887242097So the upper bound is≈1.4142011784441828Compare to m/n=239/169≈1.414201183431953Thus, m/n≈1.4142011834 versus upper bound≈1.4142011784. Therefore, m/n is indeed greater than the upper bound by approximately 0.000000005. This is a very small difference, but technically, the inequality m/n < sqrt(2)(1 -1/(4n²)) is violated here. Therefore, the problem statement as given is false. But the problem says "Prove that...", so there must be something wrong. Wait, perhaps I made an error in interpreting the problem. Let me check again. The problem states: Let m, n ∈ N and m < sqrt(2) n. Prove that m/n < sqrt(2)(1 -1/(4n²)).But for n=169 and m=239, m < sqrt(2) n holds (239 <1.41421356237*169≈239.0001), but m/n≈1.4142011834 is not less than sqrt(2)(1 -1/(4n²))≈1.4142011784. Therefore, the inequality fails. This suggests that the problem statement is incorrect. But the user is asking to prove it, so either there's a restriction missing in the problem or there's a different approach. Alternatively, maybe the original problem has a different inequality, like m/n < sqrt(2) - 1/(4n²), but the user wrote sqrt(2)(1 -1/(4n²)). Let's see the difference. If it were sqrt(2) -1/(4n²), then for n=169, the upper bound would be≈1.41421356237 -1/(4*28561)=1.41421356237 -1/114244≈1.41421356237 -0.00000875676≈1.41420480561, which is larger than m/n≈1.4142011834, so the inequality would hold. But the problem says sqrt(2)(1 -1/(4n²)), which is approximately sqrt(2) - sqrt(2)/(4n²). For n=169, this is≈1.41421356237 -1.41421356237/(4*28561)≈1.41421356237 -1.41421356237/114244≈1.41421356237 -0.0000123839≈1.4142011784, which is less than m/n. Therefore, if the problem had the inequality m/n < sqrt(2) -1/(4n²), then it would hold, but as stated, with the multiplier sqrt(2) on (1 -1/(4n²)), it's not true for n=169, m=239. Alternatively, maybe the problem has a different constraint, like m < sqrt(2) n - c for some c. But the user specified m < sqrt(2) n. Alternatively, perhaps the original problem was in a different form, and during translation or transcription, a typo occurred. Alternatively, maybe the user intended to write something else. Alternatively, maybe my entire approach is wrong, and there's a different way to prove the inequality that circumvents this issue. Let me think again. Starting over, given m < sqrt(2) n, need to show m/n < sqrt(2)(1 -1/(4n²)).Alternatively, rearrange the inequality:m/n < sqrt(2) - sqrt(2)/(4n²)So, sqrt(2) - m/n > sqrt(2)/(4n²)Multiply both sides by n:sqrt(2) n - m > sqrt(2)/(4n)So need to show that sqrt(2) n - m > sqrt(2)/(4n)Let k = sqrt(2) n - m. So need to show k > sqrt(2)/(4n)But k is the difference between sqrt(2) n and the integer m. Since m is the largest integer less than sqrt(2) n, k = {sqrt(2) n}, where {x} denotes the fractional part of x. But fractional parts of sqrt(2) n are distributed in [0,1). However, for specific n, especially those that are convergents, the fractional part can be very small. But in the case of n=169, k≈0.0001, which is less than sqrt(2)/(4n)≈0.00001238. Wait, no, k≈0.0001 is actually larger than 0.00001238. Wait, sqrt(2)/(4n) for n=169 is sqrt(2)/(4*169)=1.4142/676≈0.00209. Wait, no, wait. sqrt(2)/(4n) is 1.4142/(4*169)=1.4142/676≈0.00209. But k= sqrt(2)*n -m≈0.0001, which is much smaller than 0.00209. Therefore, k=0.0001 <0.00209. Wait, this is confusing. Let me compute sqrt(2)/(4n) for n=169:sqrt(2)/(4*169)=1.41421356237/(676)≈0.00209229334. But k=sqrt(2)*169 -239≈0.0001. So 0.0001 <0.00209229334. Therefore, in this case, k=0.0001 < sqrt(2)/(4n)=0.00209. Therefore, the inequality k > sqrt(2)/(4n) is false, which would imply that our desired result is false. But according to the problem statement, it should hold. Therefore, this suggests that either there's a mistake in the problem statement, or my approach is incorrect. Alternatively, maybe I'm missing something in the problem's constraints. Let me re-express the problem in another way. Suppose m and n are natural numbers, and m < sqrt(2) n. We need to show that m/n < sqrt(2)(1 -1/(4n²)). Let me consider rearranging the inequality:m < sqrt(2) n - sqrt(2)/(4n)But since m is an integer, this would imply that floor(sqrt(2) n) ≤ sqrt(2) n - sqrt(2)/(4n). But this is not necessarily true. For example, with n=169, sqrt(2)*169≈239.0001, so floor(sqrt(2)*169)=239. The right-hand side is sqrt(2)*169 - sqrt(2)/(4*169)≈239.0001 -0.002092≈238.998, which is less than 239. Therefore, floor(sqrt(2) n) =239 > right-hand side≈238.998. Therefore, the inequality floor(sqrt(2) n) ≤ sqrt(2) n - sqrt(2)/(4n) is false. Therefore, the problem statement's inequality does not hold for n=169, m=239. Hence, the problem as stated is incorrect. But since the user asks to prove it, perhaps there's an error in my analysis. Alternatively, maybe the original problem has an additional condition, such as n ≥2, or m and n being coprime. But even so, in the case of n=169 and m=239, they are coprime, and the inequality still fails. Alternatively, perhaps the inequality is supposed to be the other way, but that seems unlikely. Alternatively, perhaps the user made a typo and the denominator in the inequality is 2n² instead of 4n². Let me check. If it were 1 -1/(2n²), then the upper bound would be sqrt(2)(1 -1/(2n²)). For n=169, this would be≈1.41421356237*(1 -1/(2*28561))≈1.41421356237*(1 -0.0000175135)≈1.41421356237 -1.41421356237*0.0000175135≈1.41421356237 -0.00002477≈1.41418879. Then m/n=239/169≈1.4142011834 >1.41418879. So still, the inequality would fail. Alternatively, if the denominator were 8n², then 1 -1/(8n²), for n=169:≈1.41421356237*(1 -1/(8*28561))≈1.41421356237*(1 -0.000004378)≈1.41421356237 -1.41421356237*0.000004378≈1.41421356237 -0.000006196≈1.414207366, which is still less than m/n≈1.4142011834. So no, even changing the denominator doesn't fix it for this case. Alternatively, perhaps the original problem has a different exponent, like 1/(4n) instead of 1/(4n²). Let's test that. For n=169, sqrt(2)*(1 -1/(4*169))≈1.41421356237*(1 -1/676)≈1.41421356237*(0.99852071)≈1.41421356237 -1.41421356237*0.00147929≈1.41421356237 -0.002092≈1.41212156. Which is way less than m/n≈1.4142011834. So inequality would hold, but this is changing the problem. Alternatively, maybe the original problem was to prove that m/n < sqrt(2) -1/(4n²), which would avoid multiplying sqrt(2) by (1 - ...). Let's compute that. For n=169, sqrt(2) -1/(4n²)=1.41421356237 -1/(4*28561)=1.41421356237 -1/114244≈1.41421356237 -0.000008756≈1.414204806, which is greater than m/n≈1.4142011834. So the inequality would hold. But the problem as stated has sqrt(2)(1 -1/(4n²)) which is≈ sqrt(2) - sqrt(2)/(4n²), which is less than sqrt(2) -1/(4n²) since sqrt(2) >1. So the given bound is tighter. Therefore, it seems the problem statement is incorrect as the inequality does not hold for all natural numbers n. However, the user likely expects a proof, so perhaps there is a different approach that I'm missing. Let me try another angle. Suppose we start with m < sqrt(2) n. We need to show that m/n < sqrt(2)(1 -1/(4n²)). Let me assume that m is the largest integer less than sqrt(2) n. Then m = floor(sqrt(2) n). Therefore, m/n = floor(sqrt(2) n)/n. We need to show that this is less than sqrt(2)(1 -1/(4n²)). Which would imply that floor(sqrt(2) n)/n < sqrt(2) - sqrt(2)/(4n²)Rearranged: floor(sqrt(2) n) < sqrt(2) n - sqrt(2)/(4n)But floor(sqrt(2) n) = sqrt(2) n - k, where k is the fractional part {sqrt(2) n}. So we have sqrt(2) n -k < sqrt(2) n - sqrt(2)/(4n)Subtract sqrt(2) n: -k < -sqrt(2)/(4n)Multiply by -1: k > sqrt(2)/(4n)So this again reduces to showing that the fractional part of sqrt(2) n is greater than sqrt(2)/(4n). But as shown earlier, this is not true for n=169. So unless there is a lower bound on the fractional part of sqrt(2) n which is greater than sqrt(2)/(4n), but apparently, for some n, the fractional part can be smaller. However, in the specific case of n=169, the fractional part is approximately0.0001, which is about 1.0×10^-4, while sqrt(2)/(4n)≈0.00209, so yes, 0.0001 <0.00209, so the inequality fails. Therefore, unless the problem has a different condition, such as n being less than a certain value, but the problem states for all natural numbers m, n with m < sqrt(2) n. Alternatively, perhaps the original problem was to show that m/n < sqrt(2) -1/(4n²), which would be a valid inequality. Let's verify for n=169:sqrt(2) -1/(4n²)≈1.41421356237 -0.000008756≈1.414204806, which is greater than m/n≈1.4142011834. So the inequality holds. Similarly, for n=5:sqrt(2) -1/(4*25)=1.41421356237 -0.01≈1.40421356237, while m/n=7/5=1.4 <1.4042, so holds. For n=29:sqrt(2) -1/(4*841)=1.41421356237 -0.000297≈1.41391656237, which is greater than m/n=41/29≈1.413793, so holds. For n=70:sqrt(2) -1/(4*4900)=1.41421356237 -0.000051≈1.41416256237, and m/n=99/70≈1.4142857, which is greater than sqrt(2) -1/(4n²). Wait, but here m/n=99/70≈1.4142857, which is greater than sqrt(2)≈1.41421356237. But since m must be less than sqrt(2) n, m=99 is less than sqrt(2)*70≈98.9949. Wait, wait, sqrt(2)*70≈98.9949, so m must be less than that, so m=98. Then m/n=98/70=1.4. Which is less than sqrt(2) -1/(4*4900)=≈1.41421356237 -0.000051≈1.41416256. But 1.4 <1.41416256, so holds. But if m=99, which would be greater than sqrt(2)*70≈98.9949, so m=99 is not allowed. Therefore, the maximum m is 98. So in that case, m/n=98/70=1.4 <1.41416256. Therefore, if the inequality were m/n < sqrt(2) -1/(4n²), then it would hold. But the problem states m/n < sqrt(2)(1 -1/(4n²)), which is different. Alternatively, perhaps the user made a mistake in writing the inequality. Given that, perhaps the intended inequality is m/n < sqrt(2) -1/(4n²), and the factor of sqrt(2) is a typo. If that's the case, then the proof would work. Alternatively, perhaps there is a different approach to prove the original inequality, assuming that the problem is correct and my counterexample is wrong. Let me try another approach. Suppose we consider the function f(n) = sqrt(2) n - m, where m is the integer part of sqrt(2) n. We need to show that f(n) > sqrt(2)/(4n). Assume that for some n, f(n) ≤ sqrt(2)/(4n). Then:sqrt(2) n - m ≤ sqrt(2)/(4n)Multiply both sides by n:sqrt(2) n² - mn ≤ sqrt(2)/4Rearranged:sqrt(2) n² - sqrt(2)/4 ≤ mnBut mn is less than sqrt(2) n² (since m < sqrt(2) n). So:sqrt(2) n² - sqrt(2)/4 < sqrt(2) n²Which is true, but doesn't give us a contradiction. Alternatively, consider squaring both sides of the original inequality m/n < sqrt(2)(1 -1/(4n²)). So (m/n)² < 2(1 -1/(4n²))²Expand the right-hand side:2(1 -1/(2n²) +1/(16n⁴)) ≈2 -1/n² +1/(8n⁴)But m²/n² < 2 -1/n² +1/(8n⁴)Multiply both sides by n²:m² <2n² -1 +1/(8n²)But since m < sqrt(2) n, m² <2n². So m² is at most floor(2n² -1). But floor(2n² -1)=2n² -1 if 2n² is not an integer, which it never is since sqrt(2) is irrational. Therefore, m² ≤2n² -1. Therefore, we need to show that 2n² -1 <2n² -1 +1/(8n²), which is true, but this doesn't help. Alternatively, perhaps use the fact that since m < sqrt(2) n, then 2n² -m² >0. Let's consider 2n² -m². If we can show that 2n² -m² ≥1, which is true because m and n are integers and sqrt(2) is irrational, so 2n² -m² is at least 1. Then:2n² -m² ≥1Divide both sides by n²:2 - (m/n)² ≥1/n²Therefore:(m/n)² ≤2 -1/n²Take square roots (since m/n < sqrt(2)):m/n ≤ sqrt(2 -1/n²)Now, we need to compare sqrt(2 -1/n²) with sqrt(2)(1 -1/(4n²)). Let me compute sqrt(2 -1/n²). Using a binomial approximation for sqrt(2 - ε) where ε=1/n²:sqrt(2 -1/n²)=sqrt(2(1 -1/(2n²))) =sqrt(2)sqrt(1 -1/(2n²)) ≈sqrt(2)(1 -1/(4n²) -1/(32n⁴) +... )Therefore, sqrt(2 -1/n²)≈sqrt(2)(1 -1/(4n²) -1/(32n⁴))Therefore, sqrt(2 -1/n²) < sqrt(2)(1 -1/(4n²))Thus, from m/n ≤ sqrt(2 -1/n²) < sqrt(2)(1 -1/(4n²)), we can conclude that m/n < sqrt(2)(1 -1/(4n²)), which is the desired result. Wait, this seems promising. Let me go through this step by step. Since m < sqrt(2) n, squaring both sides (since both are positive) gives m² <2n². Since m and n are integers, this implies m² ≤2n² -1 (because the next integer below 2n² is 2n² -1). Therefore, 2n² -m² ≥1. Dividing both sides by n² gives:2 - (m/n)² ≥1/n²Rearranged:(m/n)² ≤2 -1/n²Taking square roots on both sides (since m/n is positive):m/n ≤sqrt(2 -1/n²)Now, expand sqrt(2 -1/n²) using a Taylor series approximation. Let me write 2 -1/n² as 2(1 -1/(2n²)). Therefore:sqrt(2(1 -1/(2n²))) =sqrt(2) * sqrt(1 -1/(2n²))Using the binomial approximation for sqrt(1 -x)≈1 -x/2 -x²/8 -..., valid for small x:sqrt(1 -1/(2n²))≈1 - (1/(2n²))/2 - (1/(2n²))²/8 + ...=1 -1/(4n²) -1/(32n⁴) +...Therefore:sqrt(2 -1/n²)=sqrt(2)*(1 -1/(4n²) -1/(32n⁴) +...)Thus:sqrt(2 -1/n²)=sqrt(2)(1 -1/(4n²)) - sqrt(2)/(32n⁴) +...Therefore, sqrt(2 -1/n²) < sqrt(2)(1 -1/(4n²))Hence, combining with the previous inequality:m/n ≤sqrt(2 -1/n²) < sqrt(2)(1 -1/(4n²))Therefore, m/n < sqrt(2)(1 -1/(4n²))Which is the desired inequality. Therefore, this approach works and proves the inequality for all natural numbers n. But wait, in the case of n=169 and m=239, we saw that m/n≈1.4142011834 and sqrt(2)(1 -1/(4n²))≈1.4142011784. So according to this proof, m/n < sqrt(2)(1 -1/(4n²)), but numerically, m/n is slightly larger. This suggests a contradiction. However, this discrepancy arises due to the approximation in the Taylor series. The expansion sqrt(1 -x)≈1 -x/2 -x²/8 is an approximation, and the higher-order terms are negative, so the actual value of sqrt(1 -x) is less than the approximation 1 -x/2. Therefore, the inequality sqrt(1 -x) <1 -x/2 holds for small x>0. Therefore, the step where we use the binomial expansion:sqrt(1 -1/(2n²)) <1 -1/(4n²)This is because the next term in the expansion is negative, so truncating after the first term gives an overestimate. Therefore, sqrt(1 -1/(2n²)) <1 -1/(4n²)Thus, multiplying by sqrt(2):sqrt(2 -1/n²) =sqrt(2) * sqrt(1 -1/(2n²)) <sqrt(2)(1 -1/(4n²))Therefore, the inequality m/n < sqrt(2)(1 -1/(4n²)) holds. But wait, in the specific case of n=169, this is contradicted by the numerical calculation. What's the resolution here? Ah, the key is that the Taylor series provides an upper bound. However, when we do the binomial expansion, we have:sqrt(1 -x) =1 -x/2 -x²/8 -x³/16 -... So truncating after the x/2 term gives an overestimate. Therefore, sqrt(1 -x) <1 -x/2. Therefore, even for x=1/(2n²), we have:sqrt(1 -1/(2n²)) <1 -1/(4n²)Therefore, sqrt(2 -1/n²)=sqrt(2) sqrt(1 -1/(2n²)) <sqrt(2)(1 -1/(4n²))Hence, the inequality m/n ≤sqrt(2 -1/n²) <sqrt(2)(1 -1/(4n²)) holds strictly. But why does the numerical calculation for n=169 show m/n exceeding the upper bound? It must be due to rounding errors or precision issues. Let me recalculate with higher precision. First, compute sqrt(2 -1/n²) for n=169:1/n²=1/28561≈0.000035012So 2 -1/n²≈2 -0.000035012≈1.999964988sqrt(1.999964988)=?Let me compute sqrt(2 -ε) where ε=0.000035012Using the expansion sqrt(2 -ε)=sqrt(2) - (ε)/(2 sqrt(2)) - (ε²)/(8 (sqrt(2))^3) +...Compute up to the second term:sqrt(2 -ε)≈sqrt(2) - ε/(2 sqrt(2))≈1.41421356237 -0.000035012/(2*1.41421356237)≈1.41421356237 -0.000035012/2.82842712474≈1.41421356237 -0.000012375≈1.414201187Compare this to m/n=239/169≈1.414201183431953So sqrt(2 -1/n²)≈1.414201187 > m/n≈1.414201183. Therefore, m/n < sqrt(2 -1/n²). Then, since sqrt(2 -1/n²) < sqrt(2)(1 -1/(4n²)), which we have shown via the binomial approximation, which holds because the higher-order terms are negative. Therefore, even though numerically the upper bound sqrt(2)(1 -1/(4n²))≈1.4142011784 seems less than m/n≈1.4142011834, it's actually due to approximation errors in the computation. In reality, the exact value of sqrt(2)(1 -1/(4n²)) is slightly less than the exact value of sqrt(2 -1/n²), which in turn is slightly greater than m/n. However, due to the precision limits of numerical calculations, when computed with finite decimal places, it might appear that m/n exceeds the upper bound, but analytically, the inequality holds. Therefore, the proof via expanding sqrt(2 -1/n²) and showing it's less than sqrt(2)(1 -1/(4n²)) is valid, and hence the original inequality holds for all natural numbers n. This resolves the apparent contradiction. Therefore, the correct proof is as follows.

✅:To prove that for natural numbers ( m, n ) with ( m < sqrt{2}n ), the inequality ( frac{m}{n} < sqrt{2}left(1 - frac{1}{4n^2}right) ) holds, we proceed as follows:Step 1: Start with the given condition ( m < sqrt{2}n ). Squaring both sides (valid since both sides are positive), we get:[m^2 < 2n^2]Since ( m ) and ( n ) are integers, ( m^2 leq 2n^2 - 1 ).Step 2: Rearrange the inequality ( 2n^2 - m^2 geq 1 ) and divide by ( n^2 ):[2 - left(frac{m}{n}right)^2 geq frac{1}{n^2}]This simplifies to:[left(frac{m}{n}right)^2 leq 2 - frac{1}{n^2}]Step 3: Take the square root of both sides (noting ( frac{m}{n} < sqrt{2} )):[frac{m}{n} leq sqrt{2 - frac{1}{n^2}}]Step 4: Expand ( sqrt{2 - frac{1}{n^2}} ) using a binomial approximation. Factor out ( sqrt{2} ):[sqrt{2 - frac{1}{n^2}} = sqrt{2} cdot sqrt{1 - frac{1}{2n^2}}]Using the approximation ( sqrt{1 - x} approx 1 - frac{x}{2} - frac{x^2}{8} ) for small ( x ), substitute ( x = frac{1}{2n^2} ):[sqrt{1 - frac{1}{2n^2}} < 1 - frac{1}{4n^2}]Multiplying by ( sqrt{2} ):[sqrt{2 - frac{1}{n^2}} < sqrt{2}left(1 - frac{1}{4n^2}right)]Step 5: Combine the results from Steps 3 and 4:[frac{m}{n} leq sqrt{2 - frac{1}{n^2}} < sqrt{2}left(1 - frac{1}{4n^2}right)]Thus:[frac{m}{n} < sqrt{2}left(1 - frac{1}{4n^2}right)]Conclusion: The inequality holds for all natural numbers ( m ) and ( n ) where ( m < sqrt{2}n ). The key step uses the binomial approximation to bound ( sqrt{2 - frac{1}{n^2}} ), ensuring the strict inequality despite potential numerical precision artifacts.Final Answerboxed{frac{m}{n} < sqrt{2}left(1 - frac{1}{4n^2}right)}

🔑:1. Given the condition ( m < sqrt{2} n ), we square both sides to obtain: [ m^2 < (sqrt{2} n)^2 ]2. Simplifying the right-hand side, we get: [ m^2 < 2n^2 ]3. Since both ( m^2 ) and ( 2n^2 ) are natural numbers, we can state that: [ m^2 leq 2n^2 - 1 ] This use of the inequality ( m^2 < 2n^2 ) is common in problems where integer solutions are involved, ensuring ( m^2 ) is still a natural number.4. Dividing both sides of the inequality by ( n^2 ), we have: [ frac{m^2}{n^2} leq 2 - frac{1}{n^2} ]5. Note that: [ frac{m^2}{n^2} leq 2 left(1 - frac{1}{2n^2}right) ]6. Taking the square root on both sides of the inequality, we obtain: [ frac{m}{n} leq sqrt{2 left(1 - frac{1}{2n^2}right)} ]7. To further approximate, using a binomial expansion or Taylor series expansion for the square root, note that: [ sqrt{1 - frac{1}{2n^2}} approx 1 - frac{1}{4n^2} + frac{1}{32n^4} text{ (truncating at the second term)} ]8. Therefore: [ sqrt{2 left(1 - frac{1}{2n^2}right)} < sqrt{2 left(1 - frac{1}{2n^2} + frac{1}{16n^4}right)} ] This inequality helps to simplify the expression within the square root.9. Hence: [ sqrt{2 left(1 - frac{1}{2n^2} + frac{1}{16n^4}right)} = sqrt{2} left(1 - frac{1}{4n^2}right) ] The approximation allows us to rewrite the inequality in a more usable form.10. Combining the results, we get: [ frac{m}{n} leq sqrt{2} left(1 - frac{1}{4n^2}right) ] Conclusion: [ boxed{frac{m}{n} < sqrt{2} left(1 - frac{1}{4n^2}right)} ]