🧠:Okay, let's see. I need to solve this system of equations. Hmm, all three equations look similar, just cyclic permutations of x, y, and z. That symmetry might be a clue. Maybe the solution has x = y = z? Let me check that first.If x = y = z, then all three equations should be the same. Let's substitute y and z with x in the first equation:x² + 7x + 2 = 2x + 4√(7x - 3)Simplify the left side: x² + 7x + 2Right side: 2x + 4√(7x - 3)Bring all terms to one side: x² + 7x + 2 - 2x - 4√(7x - 3) = 0Simplify: x² + 5x + 2 - 4√(7x - 3) = 0Hmm, this seems a bit complicated. Let me denote √(7x - 3) as t. Then 7x - 3 = t², so x = (t² + 3)/7. Substitute back into the equation:[(t² + 3)/7]² + 5*(t² + 3)/7 + 2 - 4t = 0Compute each term step by step.First term: [(t² + 3)²]/49Second term: 5*(t² + 3)/7 = (5t² + 15)/7Third term: 2Fourth term: -4tSo combining all terms:[(t² + 3)²]/49 + (5t² + 15)/7 + 2 - 4t = 0Multiply all terms by 49 to eliminate denominators:(t² + 3)² + 49*(5t² + 15)/7 + 49*2 - 49*4t = 0Simplify:(t² + 3)² + 7*(5t² + 15) + 98 - 196t = 0Compute each part:(t² + 3)² = t⁴ + 6t² + 97*(5t² +15) = 35t² + 105So adding all terms:t⁴ + 6t² + 9 + 35t² + 105 + 98 - 196t = 0Combine like terms:t⁴ + (6t² +35t²) + (9 +105 +98) -196t = 0Which is:t⁴ +41t² +212 -196t =0Arrange in descending powers:t⁴ -196t +41t² +212=0Wait, maybe rearrange terms:t⁴ +41t² -196t +212=0This is a quartic equation. Solving quartic equations is complicated. Maybe factorable?Let me see if there are rational roots using Rational Root Theorem. Possible roots are factors of 212 divided by 1: ±1, ±2, ±4, ±53, ±106, ±212.Test t=1: 1 +41 -196 +212 = 1+41=42; 42-196=-154; -154+212=58 ≠0t=2: 16 +41*4 -196*2 +212 = 16 +164 -392 +212 = (16+164)=180; (180-392)=-212; -212+212=0. Oh! t=2 is a root.So (t -2) is a factor. Let's perform polynomial division or use synthetic division.Divide t⁴ +41t² -196t +212 by (t -2). Let me use synthetic division for t=2.Coefficients: 1 (t⁴), 0 (t³), 41 (t²), -196 (t), 212 (constant)Bring down 1.Multiply by 2: 1*2=2. Add to next coefficient: 0+2=2.Multiply by 2: 2*2=4. Add to next coefficient:41 +4=45.Multiply by 2:45*2=90. Add to next coefficient: -196 +90= -106.Multiply by 2: -106*2= -212. Add to last coefficient:212 + (-212)=0. Perfect.So the quartic factors as (t -2)(t³ +2t² +45t -106). Now we need to factor the cubic.Try rational roots for the cubic: t³ +2t² +45t -106. Possible roots are ±1, ±2, ±53, etc. Test t=2: 8 +8 +90 -106= (8+8)=16; 16+90=106; 106-106=0. So t=2 is another root.Thus, cubic factors as (t -2)(t² +4t +53). So overall quartic factors as (t -2)^2(t² +4t +53). The quadratic t² +4t +53 has discriminant 16 -212 = -196, which is negative. So only real roots are t=2 (double root).Therefore, t=2. Since t = √(7x -3), so √(7x -3)=2. Square both sides:7x -3=4 =>7x=7 =>x=1.So if x=y=z=1, let's check if this satisfies the original equation.Check first equation:Left:1² +7*1 +2=1+7+2=10Right:2*1 +4√(7*1 -3)=2 +4√4=2+4*2=10. Okay, works.Similarly for the other equations. So (1,1,1) is a solution.But maybe there are other solutions where x, y, z are not all equal. Let's check.Given the symmetry, maybe another solution where variables are equal but different from 1? Wait, the previous equation when assuming x=y=z only gave x=1. So maybe only one solution in that case. But could there be solutions where variables are different?Alternatively, perhaps another set of solutions where variables take the same value but different from 1. Wait, but when we set x=y=z, we only found x=1. So unless there's a mistake in that process, perhaps that's the only solution in the symmetric case. But maybe there are asymmetric solutions.Alternatively, maybe another solution where variables are different but satisfy some relationship. Let me consider the possibility.First, note that the equations are cyclic. So perhaps if one variable is 1, others might be 1 as well. But perhaps not. Let me suppose that one variable is different.Alternatively, think about the functions involved. Each equation has a square term, linear terms, and a square root. The square roots require that the arguments inside are non-negative. So:7x -3 ≥0 ⇒x ≥3/7 ≈0.4286Similarly, y ≥3/7, z ≥3/7.Also, the square roots are 4√(7x -3), which must be real and non-negative. So variables must be at least 3/7.Another thought: perhaps use substitution. For example, express each variable in terms of another. But given the cyclic nature, it's tricky.Alternatively, subtract pairs of equations. Let me subtract the first equation from the second:(y² +7z +2) - (x² +7y +2) = (2x +4√(7y -3)) - (2z +4√(7x -3))Simplify left side: y² -x² +7z -7yRight side:2x -2z +4√(7y -3) -4√(7x -3)Factor left side: (y -x)(y +x) +7(z - y)Right side:2(x - z) +4[√(7y -3) -√(7x -3)]Hmm, this seems complicated, but maybe factor further. Let me rearrange terms:Left: (y -x)(y +x -7) +7(z - y)Wait, not sure. Let's keep as:Left: y² -x² +7(z - y)Right:2(x - z) +4[√(7y -3) -√(7x -3)]Alternatively, rearrange the left side:y² -x² = (y -x)(y +x)So:(y -x)(y +x) +7(z - y) = 2(x - z) +4[√(7y -3) -√(7x -3)]Bring all terms to left side:(y -x)(y +x) +7(z - y) -2(x - z) -4[√(7y -3) -√(7x -3)] =0Simplify:(y -x)(y +x) +7z -7y -2x +2z -4√(7y -3) +4√(7x -3)=0Combine like terms:(y -x)(y +x) + (7z +2z) + (-7y -2x) +4[√(7x -3) -√(7y -3)] =0Which is:(y -x)(y +x) +9z -7y -2x +4[√(7x -3) -√(7y -3)] =0This is getting too messy. Maybe there's another approach.Alternatively, consider substituting variables. Let me set a = √(7x -3), b = √(7y -3), c = √(7z -3). Then:7x -3 = a² ⇒x = (a² +3)/7Similarly, y = (b² +3)/7, z = (c² +3)/7Substitute into the first equation:x² +7y +2 = 2z +4aCompute x²: [(a² +3)/7]^2 = (a⁴ +6a² +9)/497y:7*(b² +3)/7 = b² +3Left side: (a⁴ +6a² +9)/49 + b² +3 +2 = (a⁴ +6a² +9)/49 + b² +5Right side:2*(c² +3)/7 +4a = (2c² +6)/7 +4aSo equation becomes:(a⁴ +6a² +9)/49 + b² +5 = (2c² +6)/7 +4aMultiply all terms by 49 to eliminate denominators:a⁴ +6a² +9 +49b² +245 = 14c² +42 +196aSimplify:a⁴ +6a² +9 +49b² +245 -14c² -42 -196a =0Combine constants:9+245-42=212So:a⁴ +6a² +49b² -14c² -196a +212 =0Similarly, for the second equation:y² +7z +2 =2x +4bFollowing similar steps:y² = [(b² +3)/7]^2 = (b⁴ +6b² +9)/497z =7*(c² +3)/7 =c² +3Left side: (b⁴ +6b² +9)/49 +c² +3 +2 = (b⁴ +6b² +9)/49 +c² +5Right side:2*(a² +3)/7 +4b = (2a² +6)/7 +4bMultiply all terms by 49:b⁴ +6b² +9 +49c² +245 =14a² +42 +196bBring all terms to left:b⁴ +6b² +9 +49c² +245 -14a² -42 -196b=0Simplify:b⁴ +6b² +49c² -14a² -196b +212 =0Third equation similarly would be:c⁴ +6c² +49a² -14b² -196c +212 =0Now, this system seems even more complicated. But maybe there's symmetry here. Since original equations are cyclic, maybe a = b = c. Let's check.If a = b = c, then substitute into first equation:a⁴ +6a² +49a² -14a² -196a +212 =0Simplify:a⁴ + (6a² +49a² -14a²) -196a +212 = a⁴ +41a² -196a +212 =0Which is the same quartic as before. So a=2, since we had t=2 earlier.Therefore, a = b = c =2. Then x = (4 +3)/7 =1, y=1, z=1. So that's the solution we found before.But perhaps there are solutions where a, b, c are different. Let's see.Alternatively, maybe other solutions where variables are different but related. Maybe a cyclic permutation where a, b, c are different but follow some pattern. However, given the symmetry, it's challenging.Alternatively, think about possible integer solutions. Suppose x, y, z are integers. Since x=1 works, maybe others? Let's check possible integers.Given x ≥3/7≈0.4286, possible integers x≥1. Suppose x=1, y=1, z=1 works. What about x=2?Check if x=2 is possible. Let's check the first equation with x=2:Left:2² +7y +2=4 +7y +2=6 +7yRight:2z +4√(14 -3)=2z +4√11≈2z +4*3.316≈2z +13.265So equation:6 +7y ≈2z +13.265 ⇒7y ≈2z +7.265But y and z need to be at least 3/7. If y and z are integers, 7y and 2z must be integers. 7.265 is ≈29/4? Not sure. Seems messy. Probably not integer solutions here.Alternatively, maybe other solutions where variables are the same but not 1. Wait, when we assumed a = b = c, the quartic only gave a=2. So maybe no other solutions in symmetric case.Alternatively, check if there are solutions where two variables are equal and the third is different. For example, suppose x = y ≠ z. Let's try that.Set x = y. Then the first two equations become:x² +7x +2 = 2z +4√(7x -3) ...(1)x² +7z +2 = 2x +4√(7x -3) ...(2)Subtract equation (2) from equation (1):(x² +7x +2) - (x² +7z +2) = 2z -2xSimplify left side:7x -7z = 2z -2x ⇒7x -7z =2z -2x ⇒7x +2x =2z +7z ⇒9x=9z ⇒x=zBut we supposed x=y and x=z, so all three variables equal. So that leads back to the symmetric solution. Thus, no solution where two variables are equal and third different.Similarly, trying other combinations would likely lead back to x=y=z.Alternatively, consider another approach. Let me assume that variables are equal in pairs but not all three. Wait, but previous step showed that if two are equal, third must be equal as well.Alternatively, consider using inequalities. Let's think about the equations. Each has a square term and a square root. Maybe applying Cauchy-Schwarz or AM-GM?For example, in the first equation:x² +7y +2 =2z +4√(7x -3)Notice that the right side has 4√(7x -3). Let me denote √(7x -3) as a, so 7x =a² +3, x=(a² +3)/7.Similarly, substitute into the equation:[(a² +3)/7]^2 +7y +2 =2z +4aBut this seems similar to the previous substitution. Maybe not helpful.Alternatively, think of each equation as relating x and z via y, etc. Maybe express z from the first equation:From first equation:x² +7y +2 -2z =4√(7x -3)So 2z =x² +7y +2 -4√(7x -3)Thus, z=(x² +7y +2 -4√(7x -3))/2Similarly, from second equation:y² +7z +2 -2x =4√(7y -3)So 7z=2x +4√(7y -3) -y² -2Thus, z=(2x +4√(7y -3) -y² -2)/7So now we have two expressions for z in terms of x and y. Set them equal:(x² +7y +2 -4√(7x -3))/2 = (2x +4√(7y -3) -y² -2)/7Multiply both sides by 14 to eliminate denominators:7(x² +7y +2 -4√(7x -3)) =2(2x +4√(7y -3) -y² -2)Expand left side:7x² +49y +14 -28√(7x -3)Right side:4x +8√(7y -3) -2y² -4Bring all terms to left side:7x² +49y +14 -28√(7x -3) -4x -8√(7y -3) +2y² +4 =0Combine like terms:7x² +2y² -4x +49y +18 -28√(7x -3) -8√(7y -3)=0This equation looks very complicated. It's a relation between x and y. Given the complexity, it's challenging to see a path forward. Maybe this approach isn't the best.Alternatively, consider the original system's symmetry. Perhaps adding all three equations?First equation: x² +7y +2 =2z +4√(7x -3)Second:y² +7z +2 =2x +4√(7y -3)Third:z² +7x +2 =2y +4√(7z -3)Add all three equations:x² + y² + z² +7y +7z +7x +6 =2z +2x +2y +4[√(7x -3) +√(7y -3) +√(7z -3)]Simplify left side:x² + y² + z² +7(x + y + z) +6Right side:2(x + y + z) +4[√(7x -3) +√(7y -3) +√(7z -3)]Bring all terms to left:x² + y² + z² +7(x + y + z) +6 -2(x + y + z) -4[√(7x -3) +√(7y -3) +√(7z -3)] =0Simplify:x² + y² + z² +5(x + y + z) +6 -4[√(7x -3) +√(7y -3) +√(7z -3)] =0Hmm, not sure if helpful. Maybe use Cauchy-Schwarz on the square roots?But not sure. Alternatively, note that each square root term is √(7x -3). If we denote these as a, b, c as before, maybe relate to variables.But this seems to be going in circles.Alternatively, consider that the symmetric solution x=y=z=1 works. Maybe it's the only solution. To verify, check if the system could have other solutions.Suppose there exists another solution where variables are different. Let's assume x >1. Then 7x -3 >4, so √(7x -3) >2. Then the right side of first equation is 2z +4*(>2)=2z +>8. Left side: x² +7y +2. If x>1, x² increases, but 7y depends on y. But variables are interdependent. It's not straightforward.Alternatively, check for possible other solutions numerically. Suppose x=2. Compute RHS of first equation: 2z +4√(14-3)=2z +4√11≈2z +13.266. LHS:4 +7y +2=6 +7y. So 6 +7y ≈2z +13.266 ⇒7y ≈2z +7.266. If z is also large, but this is speculative.Alternatively, assume that the system has only the symmetric solution. Given the symmetry and the complexity, it's plausible that (1,1,1) is the only real solution.Another approach: Check the function behavior. Let's consider the function f(t) = t² +7s +2 -2u -4√(7t -3). For the first equation, f(x) =0 when y=s and z=u. If we can show that f(t) is increasing or has only one root, maybe uniqueness follows.But this requires analysis for each variable, which is complicated.Alternatively, since we found a solution and the system is symmetric, and given the nonlinearity, it's often the case that such symmetric systems have the symmetric solution as the only real solution. Therefore, the solution is x=y=z=1.Final AnswerThe solution to the system is boxed{(1, 1, 1)}.

✅:Given the system of equations:begin{aligned}x^2 + 7y + 2 &= 2z + 4sqrt{7x - 3} y^2 + 7z + 2 &= 2x + 4sqrt{7y - 3} z^2 + 7x + 2 &= 2y + 4sqrt{7z - 3}end{aligned}We first check the possibility of a symmetric solution where (x = y = z). Substituting (x = y = z) into the first equation:[x^2 + 7x + 2 = 2x + 4sqrt{7x - 3}]Simplifying, we get:[x^2 + 5x + 2 = 4sqrt{7x - 3}]Let (t = sqrt{7x - 3}). Then (7x - 3 = t^2) and (x = frac{t^2 + 3}{7}). Substituting (x) back into the equation:[left(frac{t^2 + 3}{7}right)^2 + 5left(frac{t^2 + 3}{7}right) + 2 = 4t]Multiplying through by 49 to clear the denominators:[(t^2 + 3)^2 + 35(t^2 + 3) + 98 = 196t]Expanding and simplifying:[t^4 + 41t^2 - 196t + 212 = 0]Using the Rational Root Theorem, we find (t = 2) is a root. Factoring out ((t - 2)^2) gives:[(t - 2)^2(t^2 + 4t + 53) = 0]The quadratic (t^2 + 4t + 53) has no real roots, so the only real solution is (t = 2). Thus, (sqrt{7x - 3} = 2) implies (7x - 3 = 4) and (x = 1). Therefore, (x = y = z = 1).Verifying this solution in the original equations:- Left side: (1^2 + 7 cdot 1 + 2 = 10)- Right side: (2 cdot 1 + 4sqrt{7 cdot 1 - 3} = 2 + 4 cdot 2 = 10)All three equations are satisfied, confirming the solution. Given the symmetry and complexity of the system, it is concluded that the only real solution is:[boxed{(1, 1, 1)}]

🔑:To solve the given system of equations:begin{aligned}x^{2} + 7y + 2 &= 2z + 4sqrt{7x - 3}, y^{2} + 7z + 2 &= 2x + 4sqrt{7y - 3}, z^{2} + 7x + 2 &= 2y + 4sqrt{7z - 3},end{aligned}we add all three equations together and then simplify:1. Add the three equations:begin{aligned}(x^{2} + 7y + 2) + (y^{2} + 7z + 2) + (z^{2} + 7x + 2) &= (2z + 4sqrt{7x - 3}) + (2x + 4sqrt{7y - 3}) + (2y + 4sqrt{7z - 3}). end{aligned}2. Combine and group like terms:x^{2} + y^{2} + z^{2} + 7y + 7z + 7x + 6 = 2z + 4sqrt{7x - 3} + 2x + 4sqrt{7y - 3} + 2y + 4sqrt{7z - 3}. 3. Rearrange to match terms and factor on the right-hand side:x^{2} + y^{2} + z^{2} + 7y + 7z + 7x + 6 - (2z + 2x + 2y) = 4sqrt{7x - 3} + 4sqrt{7y - 3} + 4sqrt{7z - 3} - 6. 4. Simplify the equation by isolating terms:x^{2} + y^{2} + z^{2} + 5x + 5y + 5z + 6 = 4sqrt{7x - 3} + 4sqrt{7y - 3} + 4sqrt{7z - 3} - 6.4. To make further simplification easier, we rewrite the left-hand side completing the square by distributing and rearranging terms:(x - 1)^{2} + (y - 1)^{2} + (z - 1)^{2} + (sqrt{7x - 3} - 2)^{2} + (sqrt{7y - 3} - 2)^{2} + (sqrt{7z - 3} - 2)^{2} = 0.5. Notice that every term in the equation is a square, and thus each must be non-negative:(x - 1)^{2} geq 0,(y - 1)^{2} geq 0,(z - 1)^{2} geq 0,(sqrt{7x - 3} - 2)^{2} geq 0,(sqrt{7y - 3} - 2)^{2} geq 0,(sqrt{7z - 3} - 2)^{2} geq 0.6. Given that the sum of non-negative terms equals zero, each term must individually be zero:(x - 1)^{2} = 0 implies x = 1,(y - 1)^{2} = 0 implies y = 1,(z - 1)^{2} = 0 implies z = 1,(sqrt{7x - 3} - 2)^{2} = 0 implies sqrt{7 cdot 1 - 3} = 2 implies sqrt{4} = 2.7. Be sure to verify the pair (x, y, z) = (1, 1, 1) into the original equations to confirm:1^{2} + 7 cdot 1 + 2 = 2 cdot 1 + 4 sqrt{7 cdot 1 - 3} implies 1 + 7 + 2 = 2 + 4(2) implies 10 = 10,1^{2} + 7 cdot 1 + 2 = 2 cdot 1 + 4 sqrt{7 cdot 1 - 3} implies 1 + 7 + 2 = 2 + 4(2) implies 10 = 10,1^{2} + 7 cdot 1 + 2 = 2 cdot 1 + 4 sqrt{7 cdot 1 - 3} implies 1 + 7 + 2 = 2 + 4(2) implies 10 = 10.Hence, the solution to the system of equations is:boxed{x = y = z = 1}.

🧠:Okay, so I need to figure out how to measure exactly 4 liters of water in the five-liter container using only the three-liter and five-liter containers. Let's start by understanding the problem. We have two containers, one holds up to 3 liters and the other up to 5 liters. The goal is to get exactly 4 liters in the 5-liter container. First, I should probably think about the possible operations I can perform. The basic actions are filling a container to its full capacity, emptying a container completely, and pouring water from one container to another until either the source is empty or the destination is full. Let me try to map out the steps. Let's denote the three-liter container as A and the five-liter container as B. The initial state is both are empty: A=0, B=0. The goal is to have B=4. Maybe the classic water pouring puzzles approach applies here. Usually, these problems involve using the differences in capacities to measure out specific amounts. Since 5 and 3 are coprime (their greatest common divisor is 1), it's possible to measure any integer volume up to their sum, which is 8 liters. But here we need 4 liters in the 5-liter container.Let me try step by step. One common method is to fill the larger container first. So, fill B to 5 liters. Then pour from B into A until A is full. Since A is 3 liters, that would leave B with 5-3=2 liters. So now A=3, B=2. Then empty A. Now pour the 2 liters from B into A. Now A=2, B=0. Fill B again to 5 liters. Pour from B into A until A is full. Since A already has 2 liters, it can take 1 more liter. So pouring 1 liter from B to A would leave B with 5-1=4 liters. That's the desired amount. Let me check that again:1. Fill B to 5 (A=0, B=5)2. Pour from B to A until A is full (A=3, B=2)3. Empty A (A=0, B=2)4. Pour B's 2 liters into A (A=2, B=0)5. Fill B to 5 again (A=2, B=5)6. Pour from B to A until A is full. Since A can take 1 liter (3-2=1), pour 1 liter into A. Now B has 5-1=4 liters. (A=3, B=4)Yes, that works. So in 6 steps, we get to the desired state. Alternatively, maybe there's another way by starting with filling the smaller container first. Let's see if that approach works as well.Start by filling A (3 liters) and pouring into B. So A=0, B=3. Then fill A again (A=3, B=3). Pour into B until it's full. B can take 2 more liters (5-3=2), so we pour 2 liters from A to B, leaving A with 1 liter. Now A=1, B=5. Then empty B (A=1, B=0). Pour the 1 liter from A to B. Now A=0, B=1. Fill A again (A=3, B=1). Pour into B. B can take 4 liters, but A has 3. So pouring all 3 into B would make B=1+3=4. That works too. Let's check the steps:1. Fill A (A=3, B=0)2. Pour A into B (A=0, B=3)3. Fill A again (A=3, B=3)4. Pour from A to B until B is full. Since B can take 2 more liters, A becomes 3-2=1 (A=1, B=5)5. Empty B (A=1, B=0)6. Pour A into B (A=0, B=1)7. Fill A again (A=3, B=1)8. Pour A into B. Now B becomes 1+3=4. (A=0, B=4)That's another valid method with 8 steps. So there are multiple ways to achieve it, but the first method seems more efficient with fewer steps. Wait, the problem didn't specify minimizing the number of steps, just to measure exactly 4 liters. Either method is correct. However, the first method requires fewer steps. Let me confirm again the first method:1. Fill B (5 liters)2. Pour into A, leaving 2 in B3. Empty A4. Pour B's 2 into A5. Fill B again6. Pour into A (which needs 1) to leave 4 in B.Yes, that's 6 steps. The other approach is 8 steps. Both are correct, but maybe the first is more straightforward.Alternatively, another way to think about it is using the equation 5x + 3y = 4. We need to find integers x and y (could be positive or negative) such that this equation holds. For example, if x=2 and y=-2: 5*2 -3*2=10-6=4. That suggests filling the 5-liter container twice and pouring out two times 3 liters. But how would that translate into steps?Fill B (5), pour into A (3), leaving 2 in B. Empty A, pour the 2 into A. Fill B again (5), pour into A until full (needs 1 liter). So that gives us 4 in B. Which is the same as the first method. So this equation approach aligns with the steps.So the key is that by filling and transferring between the two containers, we can subtract multiples of their capacities to get the desired amount. Since 5 and 3 are coprime, we can get any integer value through their combinations.Another way to verify: the problem is similar to the classic 3 and 5 gallon jug problem, where the solution involves filling, transferring, and emptying in specific sequences. The method I first described is a standard solution for such puzzles.Just to ensure I haven't missed any alternative paths, let me try a different sequence. Suppose I start by filling A and pouring into B each time. Let's see:Start with A=3, B=0. Pour into B: A=0, B=3.Fill A again: A=3, B=3. Pour into B, but B can only take 2 more, so A=1, B=5.Empty B: A=1, B=0. Pour A into B: A=0, B=1.Fill A: A=3, B=1. Pour into B: A=0, B=4. That's the same as the second method.Alternatively, starting with filling A first, as in the second approach, also works. So both approaches are valid.Is there a way to do it in fewer steps? Let's see. The first method is 6 steps: fill B, pour to A, empty A, pour to A, fill B, pour to A. Wait, perhaps the steps can be counted differently. Sometimes, filling or emptying is considered a step, as well as pouring. Let's recount the first method step count:1. Fill B (step 1)2. Pour B to A (step 2)3. Empty A (step 3)4. Pour B to A (step 4)5. Fill B (step 5)6. Pour B to A (step 6)Yes, six steps. The second method takes more steps. So the first method is more efficient.Alternatively, if someone starts by filling A and then pouring into B, but that leads to the same steps as the second method. So the first method is better.Therefore, the solution is to:1. Fill the five-liter container completely.2. Pour water from the five-liter container into the three-liter one until it's full. This leaves 2 liters in the five-liter container.3. Empty the three-liter container.4. Pour the remaining 2 liters from the five-liter container into the three-liter container.5. Fill the five-liter container again.6. Pour water from the five-liter container into the three-liter container until it's full (which requires 1 liter), leaving exactly 4 liters in the five-liter container.This sequence ensures that the larger container ends up with exactly 4 liters.

✅:To measure exactly 4 liters in the five-liter container, follow these steps:1. Fill the five-liter container completely. - *State*: 3L = 0, 5L = 5.2. Pour from the five-liter into the three-liter container until the smaller one is full. - *Result*: 3L = 3, 5L = 2.3. Empty the three-liter container. - *State*: 3L = 0, 5L = 2.4. Transfer the remaining 2 liters from the five-liter to the three-liter container. - *Result*: 3L = 2, 5L = 0.5. Refill the five-liter container to its full capacity. - *State*: 3L = 2, 5L = 5.6. Pour from the five-liter container into the three-liter container until it’s full. Since the three-liter container already has 2 liters, it can only hold 1 more liter. - *Result*: 3L = 3, 5L = 4.Final State: The five-liter container now holds exactly 4 liters of water.

🔑: Problem:There is a water tap and two containers: one with a capacity of 3 liters and another with a capacity of 5 liters. How can you measure 4 liters of water in the larger container (5-liter one)?1. Fill the 5-liter container completely: The 5-liter container now has 5 liters of water. [ text{5-liter container: } 5 text{ liters} text{3-liter container: } 0 text{ liters} ]2. Pour water from the 5-liter container into the 3-liter container until it is full: After this step, [ text{5-liter container: } 5 - 3 = 2 text{ liters} text{3-liter container: } 3 text{ liters} ]3. Empty the 3-liter container: [ text{5-liter container: } 2 text{ liters} text{3-liter container: } 0 text{ liters} ]4. Pour the remaining 2 liters from the 5-liter container into the 3-liter container: [ text{5-liter container: } 0 text{ liters} text{3-liter container: } 2 text{ liters} ]5. Fill the 5-liter container again to its full capacity: [ text{5-liter container: } 5 text{ liters} text{3-liter container: } 2 text{ liters} ]6. Pour water from the 5-liter container into the 3-liter container until it is full: To fill the 3-liter container, you need 1 liter more (since it already contains 2 liters), [ text{5-liter container: } 5 - 1 = 4 text{ liters} text{3-liter container: } 3 text{ liters} ]Now, the 5-liter container has exactly 4 liters of water. Conclusion:[boxed{4 text{ liters}}]

🧠:Okay, so I need to prove that the harmonic series up to 1/n, which is 1 + 1/2 + 1/3 + ... + 1/n, is never an integer for any natural number n. Hmm, interesting. Let me think about how to approach this.First, I remember that for n=1, the sum is just 1, which is an integer. But wait, the problem says "for any natural number n," but maybe n=1 is considered a trivial case? Wait, no, n=1 gives 1, which is an integer. So maybe the problem is supposed to say for n > 1? Let me check the original problem again. It says "for any natural number n." Hmm, but that contradicts n=1. Maybe there's a typo or maybe I misread. Wait, let me confirm: The problem states "cannot be an integer for any natural number n." But 1 is an integer. So perhaps the problem is incorrect? Or maybe it's a trick question where n=1 is the only case where it's integer, but for all n >1, it's not. Maybe the original problem intended n >1? Let me assume that for n ≥ 2, the harmonic number isn't an integer. Maybe the user made a mistake in the problem statement, but I should proceed with that assumption since for n=1, it's 1. Let me check if n=2: 1 + 1/2 = 1.5, which is not integer. n=3: 1 + 0.5 + 0.333... = 1.833..., not integer. n=4: 1 + 0.5 + 0.25 + 0.333... ≈ 2.083..., still not integer. So maybe for n ≥2, it's never an integer. So the problem likely wants to show that for n ≥2, H_n is not integer. Let me proceed with that.So the goal is to show that H_n = 1 + 1/2 + 1/3 + ... + 1/n is not an integer when n ≥2.I need to find a way to demonstrate that this sum can't be an integer. How can I approach this? Maybe considering properties of the denominators, like looking at the highest power of 2 in the denominators? Because when adding fractions, the denominators need to be considered for common denominators. Let me think.Suppose we consider the harmonic series up to n. Let me denote H_n = ∑_{k=1}^n 1/k. To add these fractions, we would need a common denominator, which is the least common multiple (LCM) of the numbers from 1 to n. Let's denote L_n = LCM(1, 2, ..., n). Then, H_n can be written as a single fraction with denominator L_n. So H_n = N / L_n for some integer N. The question is whether N is divisible by L_n, which would make H_n an integer. So, if we can show that N is not divisible by L_n, then H_n is not an integer.Alternatively, maybe we can analyze the numerator modulo some number. For example, if we can show that when all terms are expressed with the common denominator L_n, the numerator N is congruent to some non-zero value modulo a prime number, which would imply that N is not divisible by L_n, hence H_n is not integer.Another idea: Bertrand's Postulate states that for any integer m >1, there exists a prime p such that m < p < 2m. Maybe this can help in finding a prime that divides the denominator but not the numerator?Alternatively, consider the highest power of 2 less than or equal to n. Let 2^m be the highest power of 2 in [1, n]. If n ≥2, then 2^m is at least 2. Maybe in the denominator, the LCM L_n has a factor of 2^m, but in the numerator, when adding all fractions, the term 1/(2^m) would contribute an odd number to the numerator when expressed over L_n. Then, the other terms would contribute even numbers, so the total numerator would be odd, hence not divisible by 2^m. But wait, let's make this precise.Suppose that in the LCM L_n, the exponent of 2 is m, where 2^m ≤n <2^{m+1}. So, 2^m is the highest power of 2 in the denominators. Then, when we write each term 1/k as (L_n /k) / L_n, so the numerator becomes sum_{k=1}^n (L_n /k). Let's denote each term as L_n /k. The total numerator is N = sum_{k=1}^n (L_n /k). Then, if we can show that N is odd, while L_n is divisible by 2^m, then N / L_n cannot be an integer, since N is odd and L_n has at least one factor of 2. But wait, 2^m divides L_n, so if N is not divisible by 2^m, then N / L_n is not integer. But how do we know N is not divisible by 2^m?Alternatively, consider that only one term in the sum N has a denominator with 2^m. That term is 1/(2^m). So when we compute L_n / (2^m), this term would be L_n / (2^m). Since L_n is divisible by 2^m, L_n / (2^m) is an integer, and in fact, this integer is odd. Because L_n is LCM(1, 2, ..., n), which includes 2^m, but not 2^{m+1}, so when we divide by 2^m, the result would be an integer that is not divisible by 2. Therefore, L_n / (2^m) is odd. All other terms L_n /k for k ≠2^m: For k not divisible by 2^m, L_n /k is divisible by 2, since k has fewer factors of 2. For example, if k is even but not equal to 2^m, then k has at least one factor of 2 but less than m, so L_n /k would still have at least one factor of 2. Similarly, if k is odd, then L_n is divisible by 2^m, so L_n /k is divisible by 2^m, which is even. Therefore, all terms except the term corresponding to 2^m are even. The term corresponding to 2^m is odd. Therefore, the sum N is equal to even + even + ... + odd = odd. Therefore, N is odd. However, L_n is divisible by 2^m, which is at least 2. Therefore, N / L_n is a fraction with an odd numerator and even denominator. Therefore, the fraction cannot be an integer, since an even denominator cannot divide an odd numerator. Thus, H_n is not an integer.Wait, this seems like a solid approach. Let me verify this step by step.First, let m be the integer such that 2^m is the highest power of 2 less than or equal to n. So 2^m ≤n <2^{m+1}. Then, in the LCM L_n of 1 to n, the exponent of 2 is m. So L_n is divisible by 2^m but not by 2^{m+1}.When we write each term 1/k as (L_n /k)/L_n, then the numerator is sum_{k=1}^n (L_n /k). Let's call this sum N. Then H_n = N / L_n.Now, consider each term L_n /k. For k =2^m, L_n /k = L_n /2^m. Since L_n is divisible by 2^m, this is an integer. Moreover, since L_n is the LCM of numbers up to n, and 2^m is the highest power of 2 in L_n, then L_n /2^m is not divisible by 2, because otherwise L_n would be divisible by 2^{m+1}, which contradicts the definition of m. Therefore, L_n /2^m is odd.For other values of k, i.e., k ≠2^m, let's see:If k is not divisible by 2^m, then k has fewer factors of 2. Therefore, L_n /k is divisible by 2. Because L_n has exactly m factors of 2, and k has t < m factors of 2. Therefore, L_n /k has m - t ≥1 factors of 2, so it's even.If k is divisible by 2^m, but since 2^m is the highest power of 2 in L_n, the only such k is 2^m itself. Because if k were a multiple of 2^m, like 2^m * l where l ≥2, then 2^{m+1} would divide k, but since k ≤n <2^{m+1}, l must be 1. Hence, the only multiple of 2^m in 1 to n is 2^m itself. Therefore, only the term k=2^m has L_n /k odd, and all other terms have L_n /k even.Thus, the sum N = sum_{k=1}^n (L_n /k) = even + even + ... + odd = odd + even * (n-1). Since even * (n-1) is even, and odd + even is odd. Therefore, N is odd. However, L_n is divisible by 2^m, which is at least 2. Therefore, the denominator L_n has at least one factor of 2, and the numerator N is odd. Therefore, the fraction N / L_n cannot be an integer because an odd number cannot be divided by 2 to give an integer. Therefore, H_n is not an integer.This seems to hold for n ≥2. For n=1, L_1=1, N=1, so H_1=1 which is integer. So the original problem statement might have an exception for n=1, but as per the problem, maybe they consider n ≥2. Therefore, the harmonic number H_n is not integer for n ≥2.Let me check this with some examples.Take n=2: L_2=2. Then N = 2/1 + 2/2 = 2 +1=3. So H_2=3/2=1.5, not integer. Correct.n=3: L_3=6. N=6/1 +6/2 +6/3=6 +3 +2=11. 11/6≈1.833..., not integer. Correct.n=4: L_4=12. N=12/1 +12/2 +12/3 +12/4=12 +6 +4 +3=25. 25/12≈2.083..., not integer. Correct.n=5: L_5=60. N=60 +30 +20 +15 +12=137. 137/60≈2.283..., not integer. Indeed.n=6: L_6=60. N=60 +30 +20 +15 +12 +10=147. 147/60=2.45, not integer.Wait, but in n=4, L_n=12. The highest power of 2 less than or equal to 4 is 4=2^2. Then, L_n /4=3, which is odd. All other terms: L_n /1=12 (even), L_n /2=6 (even), L_n /3=4 (even). Then sum is 12 +6 +4 +3=25, which is odd. 25/12 is not integer. Correct.Similarly, n=6: highest power of 2 is 4 (since 2^2=4 ≤6 <8=2^3). Then L_n=60, which includes 4 as a factor. L_n /4=15, which is odd. The other terms:L_n /1=60 (even)L_n /2=30 (even)L_n /3=20 (even)L_n /4=15 (odd)L_n /5=12 (even)L_n /6=10 (even)Sum: 60 +30 +20 +15 +12 +10=147, which is odd. 147/60=2.45, not integer. Correct.So this approach seems valid. Therefore, the key idea is that in the sum N, only the term corresponding to the highest power of 2 in the denominators contributes an odd number, and all others contribute even numbers. Hence, the total numerator is odd, and the denominator is even, making the fraction irreducible with an even denominator, thus not an integer.Therefore, this proves that for n ≥2, H_n is not an integer. For n=1, it's 1, which is integer, but the problem statement probably had a typo or intended n ≥2. Assuming that, the proof holds.Another way to think about this is using Bertrand's Postulate. For n ≥2, there exists a prime p such that n/2 <p ≤n. Then, in the harmonic series, the term 1/p would have a denominator that is a prime not dividing any other denominators except p itself. So when we take the LCM of denominators, p would appear only once. Then, when expressing all fractions with denominator L_n, the term 1/p would contribute L_n /p to the numerator. Since p is a prime and p >n/2, it can't divide any other term's denominator except p itself. Hence, L_n is divisible by p but not by p^2. Then, in the numerator, the term L_n /p is divisible by p^{0}, since L_n has exactly one factor of p. Therefore, when we sum all terms, the numerator would have exactly one term (from 1/p) that is not divisible by p, and all others divisible by p. Hence, the total numerator would be congruent to L_n /p modulo p. Since L_n is divisible by p, L_n /p is congruent to (k*p)/p =k modulo p, where k is not divisible by p (since L_n isn't divisible by p^2). Therefore, the total numerator is congruent to k modulo p, which is not zero. Hence, the numerator is not divisible by p, but the denominator is divisible by p. Therefore, the fraction cannot be an integer.Wait, but this approach might require more steps. Let me see. If we have a prime p such that n/2 <p ≤n, which exists by Bertrand's Postulate for n ≥2. Then, in the harmonic series, the term 1/p is present. The LCM L_n is divisible by p, but not by p^2 because p ≤n and 2p >n, so there's no multiple of p^2 ≤n. Hence, in L_n, the exponent of p is 1. Therefore, when we write each term 1/k as (L_n /k)/L_n, the numerator term for k=p is L_n /p, which is an integer not divisible by p (since L_n has exactly one factor of p). For all other terms where k≠p, since k ≤n <2p, and p is a prime, k cannot be a multiple of p unless k=p. Hence, for k≠p, L_n /k is divisible by p if and only if k is a multiple of p, but since k ≠p and k <2p (as p >n/2), the only multiple is p itself. Therefore, all other terms L_n /k are divisible by p except for the term k=p. Therefore, the total numerator N = sum_{k=1}^n (L_n /k) is equal to (multiple of p) + (L_n /p). Since L_n /p is not divisible by p, the entire sum N is congruent to L_n /p modulo p. Since L_n is divisible by p, L_n /p is congruent to some integer m modulo p, and since L_n /p is not divisible by p, m ≠0 modulo p. Therefore, N ≡ m ≠0 mod p. Hence, N is not divisible by p, but L_n is divisible by p. Therefore, the fraction N / L_n cannot be an integer, as p divides the denominator but not the numerator. Therefore, H_n is not an integer.This is another approach using Bertrand's Postulate. So both methods work. The first method uses the highest power of 2, the second uses Bertrand's to find a prime p in (n/2, n]. Both lead to the conclusion that H_n is not an integer for n ≥2.Which proof is better? The first one is more elementary, not requiring Bertrand's Postulate, which might be more advanced. The second one uses Bertrand's, which is a deeper result but might cover more cases or give a different perspective.But since the problem is likely expecting an elementary proof, the first approach with the highest power of 2 is preferable.Therefore, the key steps are:1. Let 2^m be the highest power of 2 ≤n.2. When expressing H_n as N / L_n, where L_n is LCM(1,2,...,n), the term 1/2^m contributes an odd number to the numerator N, while all other terms contribute even numbers.3. Therefore, N is odd, and L_n is even.4. Hence, N / L_n cannot be an integer.This completes the proof.

✅:To prove that the harmonic number ( H_n = 1 + frac{1}{2} + frac{1}{3} + ldots + frac{1}{n} ) cannot be an integer for any natural number ( n geq 2 ), we analyze the structure of the sum through its least common multiple (LCM) representation.Key Steps:1. Highest Power of 2: Let ( 2^m ) be the highest power of 2 such that ( 2^m leq n ). For ( n geq 2 ), ( m geq 1 ).2. LCM Denominator: Let ( L_n = text{LCM}(1, 2, ldots, n) ). The exponent of 2 in ( L_n ) is ( m ), so ( L_n ) is divisible by ( 2^m ) but not ( 2^{m+1} ).3. Numerator Analysis: Express ( H_n ) as ( frac{N}{L_n} ), where ( N = sum_{k=1}^n frac{L_n}{k} ). - For ( k = 2^m ), ( frac{L_n}{2^m} ) is odd (since ( L_n ) contains ( 2^m ), but no higher power). - For ( k neq 2^m ), ( frac{L_n}{k} ) is even (as ( k ) has fewer factors of 2 than ( L_n )). 4. Parity of the Numerator: The sum ( N ) consists of one odd term (( frac{L_n}{2^m} )) and all other terms even. Thus, ( N ) is odd.5. Conclusion: Since ( N ) is odd and ( L_n ) is even, ( frac{N}{L_n} ) cannot simplify to an integer. Hence, ( H_n ) is not an integer for ( n geq 2 ).Final Answer: For any natural number ( n geq 2 ), the harmonic number ( H_n ) cannot be an integer. This is proven by analyzing the parity of the numerator when expressed over the least common multiple denominator, showing it must remain a fraction with an odd numerator and even denominator. Thus, The harmonic series ( 1 + frac{1}{2} + frac{1}{3} + ldots + frac{1}{n} ) is never an integer for ( n geq 2 ).

🔑:To prove that the number [ 1 + frac{1}{2} + frac{1}{3} + ldots + frac{1}{n} ]cannot be an integer for any natural number ( n ), we proceed as follows:1. Common Denominator Approach: - Consider the harmonic sum ( H_n = 1 + frac{1}{2} + frac{1}{3} + ldots + frac{1}{n} ). - It is well known that the least common multiple (LCM) of denominators 1, 2, 3, ..., ( n ) is a very large number (for instance, for ( n=9 ), LCM is 2520). Let's denote this by ( L ). - The sum can be rewritten with this common denominator: [ H_n = frac{L}{L} left( 1 + frac{1}{2} + frac{1}{3} + ldots + frac{1}{n} right) = frac{S}{L} ]2. Odd and Even Denominators: - Specifically, we focus on the highest power of 2 present in the denominators. Suppose among the fractions, there is a term with the highest power of 2 in its denominator, say ( frac{1}{k cdot 2^t} ). - When all fractions are converted to this common denominator ( k cdot 2^t ), the additional multipliers needed for other fractions would typically introduce odd factors or even factors containing 2.3. Parity Analysis in Numerator: - We will show that the sum ( S ) in the numerator of ( frac{S}{L} ) contains both even and odd terms: - Most terms are even in ( S ) due to conversion of fractions with higher powers of 2 being canceled out. - However, there remains at least one term that's specifically odd, making the entire numerator ( S ) odd.4. Odd Numerator and Even Denominator: - We conclude ( S ) is odd because an odd plus any mix of evens remains odd. - ( L ) is even since it's a product of multiple denominators inclusive of the factor 2. - An odd number divided by an even number cannot result in an integer.Thus, the harmonic number ( H_n ) cannot be an integer for any natural ( n ). Conclusion:[ boxed{H_n text{ cannot be an integer.}} ]

🧠:Okay, let's see. So, we have a triangle ABC with a point P inside it. The vector from A to P is given as (1/3) the vector AB plus (1/4) the vector AC. We need to find the ratios of the areas of the triangles PBC, PCA, and PAB, which are S1, S2, S3 respectively.First, I need to recall how vectors can be used to determine areas in a triangle. Maybe barycentric coordinates? Or perhaps using determinants since area can be related to the cross product of vectors. Hmm.Given the vector equation: AP = (1/3)AB + (1/4)AC. Let me try to express this in terms of position vectors. Let’s assume point A is the origin to simplify calculations. If A is the origin, then the position vectors of points B and C are just vectors AB and AC, right?Wait, if A is the origin, then the coordinates of B and C can be represented as vectors b and c. Then point P is given by the vector p = (1/3)b + (1/4)c. So, P is located at (1/3, 1/4) in the coordinate system where A is the origin, and the basis vectors are AB and AC. That might make things easier.But how does this help with the areas? Maybe I need to express the areas in terms of the coordinates of P. Since areas in triangles can be related to the ratios of the segments or using determinants.Alternatively, perhaps mass point geometry could be useful here. If AP is expressed as a combination of AB and AC, maybe we can assign masses at the vertices to balance the point P.Wait, another approach: using affine coordinates. In the affine plane, any point inside the triangle can be expressed as a combination of the vertices with coefficients that sum to 1. But in this case, the expression given is AP = (1/3)AB + (1/4)AC. Let me check the coefficients. If we consider AP as a vector starting at A, then the coefficients of AB and AC are 1/3 and 1/4. But do these coefficients sum to something? 1/3 + 1/4 = 7/12. Hmm, not 1. So maybe this isn't barycentric coordinates.Wait, barycentric coordinates typically express a point as a combination of the vertices with coefficients adding to 1. But here, the vector AP is given in terms of vectors AB and AC. So maybe we need to convert this into barycentric coordinates.Alternatively, let's consider that in vector terms, the position vector of P (with A as origin) is p = (1/3)b + (1/4)c. Therefore, the coordinates of P in terms of the basis vectors b and c are (1/3, 1/4). But in barycentric coordinates, we need weights that sum to 1. Let me see.Wait, barycentric coordinates with respect to triangle ABC would express P as αA + βB + γC where α + β + γ = 1. But since we're taking A as the origin, maybe this complicates things. Alternatively, if we write the position vector of P as a combination of vectors AB and AC, then that's equivalent to coordinates in the affine plane based at A with basis vectors AB and AC.Alternatively, maybe it's easier to work with areas directly. The areas S1, S2, S3 are the areas of the three smaller triangles formed by P inside ABC. The sum of these areas should be equal to the area of ABC. So, S1 + S2 + S3 = [ABC].Our goal is to find S1:S2:S3. Let's denote the area of ABC as T. Then S1 = [PBC], S2 = [PCA], S3 = [PAB]. So, T = S1 + S2 + S3.Since P is defined in terms of vectors from A, maybe we can express the areas in terms of the coefficients in the vector equation. Let me think.If we consider the position vector of P as p = (1/3)b + (1/4)c, then in the coordinate system where A is the origin, B is (1,0) and C is (0,1), but scaled by vectors AB and AC. Wait, maybe parametrize the triangle.Alternatively, let's use linear algebra. Let’s set coordinate system with A at (0,0), B at (3,0), and C at (0,4). Then, vector AB is (3,0), vector AC is (0,4). Then point P is given by AP = (1/3)AB + (1/4)AC = (1/3)(3,0) + (1/4)(0,4) = (1,0) + (0,1) = (1,1). So in this coordinate system, P is at (1,1). Hmm, interesting. Then, coordinates are A(0,0), B(3,0), C(0,4), P(1,1).Wait, that seems manageable. Then, we can compute the areas S1, S2, S3 directly. Let me try that.First, compute coordinates of all points:A: (0,0)B: (3,0)C: (0,4)P: (1,1)Now, let's compute the areas of the three triangles.1. S1: Area of triangle PBC.Points: P(1,1), B(3,0), C(0,4)Using the formula for the area of a triangle given coordinates:Area = (1/2)| (x_P(y_B - y_C) + x_B(y_C - y_P) + x_C(y_P - y_B) ) |Plugging in:= (1/2)| 1*(0 - 4) + 3*(4 - 1) + 0*(1 - 0) |= (1/2)| 1*(-4) + 3*(3) + 0*(1) |= (1/2)| -4 + 9 + 0 | = (1/2)|5| = 5/22. S2: Area of triangle PCA.Points: P(1,1), C(0,4), A(0,0)Area = (1/2)|x_P(y_C - y_A) + x_C(y_A - y_P) + x_A(y_P - y_C)|= (1/2)|1*(4 - 0) + 0*(0 - 1) + 0*(1 - 4)|= (1/2)|4 + 0 + 0| = (1/2)*4 = 23. S3: Area of triangle PAB.Points: P(1,1), A(0,0), B(3,0)Area = (1/2)|x_P(y_A - y_B) + x_A(y_B - y_P) + x_B(y_P - y_A)|= (1/2)|1*(0 - 0) + 0*(0 - 1) + 3*(1 - 0)|= (1/2)|0 + 0 + 3| = (1/2)*3 = 3/2So, S1 = 5/2, S2 = 2, S3 = 3/2.Now, let's express these as ratios. To convert them into integer ratios, we can multiply each by 2 to eliminate denominators:S1:5/2 *2 =5S2:2 *2=4S3:3/2 *2=3So the ratio S1:S2:S3 is 5:4:3.Wait, but let me check if this coordinate system is valid. Because in the problem, the vector AP is given as (1/3)AB + (1/4)AC. By choosing AB as (3,0) and AC as (0,4), we made it so that (1/3)AB is (1,0) and (1/4)AC is (0,1), so their sum is (1,1), which is point P. So this coordinate system is consistent with the problem's vector equation.But in reality, the actual lengths of AB and AC could be anything, but by choosing AB and AC as vectors (3,0) and (0,4), we're effectively scaling the coordinate system so that the coefficients 1/3 and 1/4 correspond to unit vectors. This should be okay because areas are relative, so the ratios should remain the same regardless of scaling.Wait, but let's verify the area of the entire triangle ABC in this coordinate system. The area of ABC with coordinates A(0,0), B(3,0), C(0,4) is (1/2)*base*height = (1/2)*3*4 = 6. Then S1 + S2 + S3 = 5/2 + 2 + 3/2 = (5/2 + 3/2) + 2 = 4 + 2 = 6, which matches. So the areas add up correctly.Therefore, the ratio S1:S2:S3 is 5:4:3. But wait, let me think again. The answer seems straightforward once we assign coordinates, but is there a way to do this without coordinates, maybe using vector ratios or barycentric coordinates?Alternatively, let's think in terms of mass point or area ratios through vectors.Given that AP = (1/3)AB + (1/4)AC. Let's express this as AP = (1/3)(B - A) + (1/4)(C - A). Since we can take A as the origin, as before, this simplifies to P = (1/3)B + (1/4)C.But in barycentric coordinates, if we express P as uA + vB + wC, then u + v + w = 1. However, here, since A is the origin, the coordinates are in terms of vectors from A, so maybe it's affine coordinates. Alternatively, perhaps we can relate the coefficients to the areas.Alternatively, in terms of affine combinations, we can write P = (1 - 1/3 - 1/4)A + 1/3 B + 1/4 C. Wait, but 1 - 1/3 -1/4 = 1 - 7/12 = 5/12. So barycentric coordinates would be (5/12, 1/3, 1/4). But barycentric coordinates (α, β, γ) correspond to weights such that α + β + γ = 1. So here, the weights are 5/12 for A, 1/3 for B, 1/4 for C. Then, the areas are proportional to the weights of the opposite vertices? Wait, maybe not exactly.In barycentric coordinates, the weights correspond to the ratios of areas. Specifically, the weight at A is proportional to the area of triangle PBC, the weight at B to the area of PCA, and the weight at C to the area of PAB. Wait, is that correct?Wait, in barycentric coordinates, the coordinates (α, β, γ) correspond to the normalized areas such that:α = [PBC]/[ABC]β = [PCA]/[ABC]γ = [PAB]/[ABC]But in this case, the barycentric coordinates are given as (5/12, 1/3, 1/4). Let me check:If α = 5/12, β = 1/3 = 4/12, γ = 1/4 = 3/12. Then the ratios α:β:γ = 5:4:3, which correspond to the areas [PBC]:[PCA]:[PAB]. So that matches our previous result.Therefore, the ratio S1:S2:S3 is 5:4:3.So whether by coordinate calculation or barycentric coordinates, we arrive at the same ratio. Therefore, the answer should be 5:4:3.But let me verify once more with another approach to ensure no mistakes.Alternative approach: Using the concept that in a triangle, if a point P divides the sides in certain ratios, the areas can be found by those ratios. However, here P is defined in terms of vectors from A, so maybe we can use the ratio of the coefficients in the vector expression.Given AP = (1/3)AB + (1/4)AC. Let's consider the ratios along AB and AC. The coefficients are 1/3 and 1/4. So, if we consider moving 1/3 of the way from A to B and 1/4 of the way from A to C, the point P is determined by those two movements.In terms of areas, the area ratios can be related to these coefficients. But how?Wait, perhaps using the formula for the area of a triangle in terms of base and height. If we consider AP as a combination of AB and AC, then the heights corresponding to these bases would affect the areas.Alternatively, if we take AB as a base, the height from C to AB is the same as the height from P to AB in triangle PAB. Wait, maybe not directly.Alternatively, think of the area ratios as determined by the position of P relative to the sides. For example, the area of PBC compared to ABC would depend on the distance from P to BC compared to the distance from A to BC. Similarly for the other areas.But since BC is opposite to A, the distance from P to BC can be found by the ratio in barycentric coordinates, which we had as α = 5/12. Therefore, the area [PBC] = α[ABC] = (5/12)T, and similarly [PCA] = βT = (1/3)T, [PAB] = γT = (1/4)T. Wait, but 5/12 + 1/3 +1/4 = 5/12 +4/12 +3/12=12/12=1, so that's correct.But then the ratio [PBC]:[PCA]:[PAB] = 5/12 : 1/3 : 1/4 = 5/12 :4/12 :3/12=5:4:3. So that's the same result.Therefore, regardless of the method, the ratio is 5:4:3. So I think this is the correct answer.But just to be thorough, let's do another check with vectors.Suppose we don't fix coordinates but use vectors.Let’s denote AB as vector b and AC as vector c. Then AP = (1/3)b + (1/4)c.Then, the position vector of P is p = A + (1/3)b + (1/4)c. Since A is the origin, p = (1/3)b + (1/4)c.The area ratios can be found by looking at the coefficients in the barycentric coordinates. As we saw earlier, barycentric coordinates (α, β, γ) correspond to the weights such that α + β + γ = 1. Here, if we express p as αA + βB + γC, since A is the origin, p = βB + γC. Therefore, β = 1/3, γ =1/4, and α = 1 - β - γ = 1 -1/3 -1/4 = 12/12 -4/12 -3/12=5/12. Therefore, barycentric coordinates are (5/12,1/3,1/4), so the area ratios are 5:4:3.Therefore, this confirms again the same result.Thus, after multiple methods and checks, the ratio S1:S2:S3 is 5:4:3.Final AnswerThe ratio ( S_{1} : S_{2} : S_{3} ) is boxed{5 : 4 : 3}.

✅:Given that point ( P ) is inside ( triangle ABC ) and satisfiesoverrightarrow{AP} = frac{1}{3} overrightarrow{AB} + frac{1}{4} overrightarrow{AC},we need to determine the ratio of the areas ( S_{1} ), ( S_{2} ), and ( S_{3} ) of triangles ( triangle PBC ), ( triangle PCA ), and ( triangle PAB ) respectively.1. Position Vector of ( P ): - Assume ( A ) is the origin. Then the position vectors of ( B ) and ( C ) are ( overrightarrow{AB} ) and ( overrightarrow{AC} ). - The position vector of ( P ) is given by: overrightarrow{AP} = frac{1}{3} overrightarrow{AB} + frac{1}{4} overrightarrow{AC} 2. Coordinate System: - Set coordinates with ( A ) at ( (0,0) ), ( B ) at ( (3,0) ), and ( C ) at ( (0,4) ). This scaling simplifies calculations. - Point ( P ) has coordinates ( (1,1) ) derived from the vector equation.3. Area Calculations: - Area of ( triangle PBC ): - Points: ( P(1,1) ), ( B(3,0) ), ( C(0,4) ) - Area ( S_{1} = frac{5}{2} ) - Area of ( triangle PCA ): - Points: ( P(1,1) ), ( C(0,4) ), ( A(0,0) ) - Area ( S_{2} = 2 ) - Area of ( triangle PAB ): - Points: ( P(1,1) ), ( A(0,0) ), ( B(3,0) ) - Area ( S_{3} = frac{3}{2} )4. Verification: - Total area of ( triangle ABC ) is 6, which matches the sum ( S_{1} + S_{2} + S_{3} = frac{5}{2} + 2 + frac{3}{2} = 6 ).5. Barycentric Coordinates: - Barycentric coordinates of ( P ) are ( left( frac{5}{12}, frac{1}{3}, frac{1}{4} right) ), corresponding to the area ratios ( 5:4:3 ).Thus, the ratio ( S_{1} : S_{2} : S_{3} ) is boxed{5 : 4 : 3}.

🔑:We are given that point ( P ) lies inside the triangle ( triangle ABC ) such that [overrightarrow{AP} = frac{1}{3} overrightarrow{AB} + frac{1}{4} overrightarrow{AC}.]Let's start from the given equation and rewrite it in terms of vectors involving point ( P ):[overrightarrow{AP} = frac{1}{3} overrightarrow{AB} + frac{1}{4} overrightarrow{AC}.]First, we express (overrightarrow{AB}) and (overrightarrow{AC}) using vectors involving (P):[overrightarrow{AB} = overrightarrow{PB} - overrightarrow{PA},][overrightarrow{AC} = overrightarrow{PC} - overrightarrow{PA}.]Substituting these into the given equation:[overrightarrow{AP} = frac{1}{3} (overrightarrow{PB} - overrightarrow{PA}) + frac{1}{4} (overrightarrow{PC} - overrightarrow{PA}).]Combine like terms:[overrightarrow{AP} = frac{1}{3} overrightarrow{PB} + frac{1}{4} overrightarrow{PC} - left(frac{1}{3} + frac{1}{4}right) overrightarrow{PA}.]Calculating the coefficient of (overrightarrow{PA}):[frac{1}{3} + frac{1}{4} = frac{4}{12} + frac{3}{12} = frac{7}{12}.]Hence,[overrightarrow{AP} = frac{1}{3} overrightarrow{PB} + frac{1}{4} overrightarrow{PC} - frac{7}{12} overrightarrow{PA}.]To isolate (overrightarrow{PA}), multiply through by 12 to clear the fractions:[12 overrightarrow{AP} = 4 overrightarrow{PB} + 3 overrightarrow{PC} - 7 overrightarrow{PA}.]Bringing all terms involving (overrightarrow{PA}) to one side:[12 overrightarrow{AP} + 7 overrightarrow{PA} = 4 overrightarrow{PB} + 3 overrightarrow{PC}.]Factor out (overrightarrow{PA}) on the left side:[19 overrightarrow{PA} = 4 overrightarrow{PB} + 3 overrightarrow{PC}.]Dividing through by 19:[overrightarrow{PA} = frac{4}{19} overrightarrow{PB} + frac{3}{19} overrightarrow{PC}.]Rewriting the expression in terms of (overrightarrow{P}):[5 overrightarrow{PA} + 4 overrightarrow{PB} + 3 overrightarrow{PC} = mathbf{0}.]Now, let's determine the area ratios of triangles ( triangle PBC ), ( triangle PCA ), and ( triangle PAB ).Observe that ( overrightarrow{AP} = overrightarrow{PA} ). Given the derived equation (5 overrightarrow{PA} + 4 overrightarrow{PB} + 3 overrightarrow{PC} = mathbf{0}), the coefficients 5, 4, and 3 represent the relative weights (or leverage) of each area. These coefficients are proportional to the areas of the corresponding triangles.Thus, the ratios of the areas of the triangles are:[S_1: S_2: S_3 = 5: 4: 3.]Conclusion:[boxed{5: 4: 3}]

🧠:Okay, let me try to solve this equation step by step. The equation given is:(log_{2} 3 + 2 log_{4} x = x^{frac{log_{9} 16}{log_{3} x}}).Hmm, this looks a bit complicated with logarithms and exponents. I need to simplify both sides and see if I can find a value for x that satisfies the equation. Let me start by recalling some logarithmic identities and properties that might help here.First, I remember that (log_{a} b = frac{ln b}{ln a}) if I need to change the base, but maybe I can manipulate the terms without converting to natural logs right away. Let's look at each term separately.Starting with the left side: (log_{2} 3 + 2 log_{4} x). The second term has base 4, which is 2 squared. Maybe I can express (log_{4} x) in terms of base 2. Since (log_{4} x = frac{log_{2} x}{log_{2} 4} = frac{log_{2} x}{2}). So, 2 times (log_{4} x) would be 2 * ( (frac{log_{2} x}{2}) ) = (log_{2} x). Therefore, the left side simplifies to (log_{2} 3 + log_{2} x). Using the logarithm addition rule, (log_{a} b + log_{a} c = log_{a} (bc)), so the left side becomes (log_{2} (3x)). So now the equation is:(log_{2} (3x) = x^{frac{log_{9} 16}{log_{3} x}}).Okay, let's look at the exponent on the right side: (frac{log_{9} 16}{log_{3} x}). Let me simplify both the numerator and the denominator.Starting with the numerator: (log_{9} 16). Since 9 is 3 squared and 16 is 4 squared or 2 to the 4th. Maybe express (log_{9} 16) in terms of base 3 or base 2. Let's try base 3 first. (log_{9} 16 = frac{log_{3} 16}{log_{3} 9} = frac{log_{3} 16}{2}). Now, (log_{3} 16) can be written as (log_{3} 2^4 = 4 log_{3} 2). So, the numerator becomes (frac{4 log_{3} 2}{2} = 2 log_{3} 2).So, the exponent is (frac{2 log_{3} 2}{log_{3} x}). This can be rewritten using logarithm rules. Remember that (frac{log_{a} b}{log_{a} c} = log_{c} b). So here, (frac{log_{3} 2}{log_{3} x} = log_{x} 2), so the exponent becomes 2 * (log_{x} 2) = (2 log_{x} 2).Alternatively, using change of base formula again: (log_{x} 2 = frac{log_{3} 2}{log_{3} x}), which is what we had before. Either way, the exponent simplifies to (2 log_{x} 2). Let me note that exponent rules: (x^{a cdot b} = (x^a)^b), but not sure if that helps here.Wait, the exponent is (2 log_{x} 2). Let me recall that (a^{log_{b} c} = c^{log_{b} a}). Maybe this identity can help. Let's see:The right side is (x^{2 log_{x} 2}). Applying the identity: (x^{log_{x} 2}) would be 2, right? Because (a^{log_{a} b} = b). So if I have (x^{log_{x} 2}) that's 2. Therefore, (x^{2 log_{x} 2} = (x^{log_{x} 2})^2 = 2^2 = 4). Wait, so does that mean the entire right side simplifies to 4?Wait, let me verify that again. Let me take the exponent: (2 log_{x} 2). Then (x^{2 log_{x} 2}) can be written as ( (x^{log_{x} 2})^2 ). Since (x^{log_{x} 2} = 2), then this becomes (2^2 = 4). So yes, the right side simplifies to 4. So the equation now is:(log_{2} (3x) = 4).So solving this equation would give me the value of x. Let me check that again because this seems too straightforward, but maybe it's correct. Let me go through the steps again to ensure I didn't make a mistake.Starting with the right side: (frac{log_{9} 16}{log_{3} x}). Converted (log_{9} 16) to (2 log_{3} 2), then divided by (log_{3} x) gives (2 log_{3} 2 / log_{3} x = 2 log_{x} 2). Then exponent of x with that is (x^{2 log_{x} 2} = (x^{log_{x} 2})^2 = 2^2 = 4). That seems correct.So the equation reduces to (log_{2} (3x) = 4). To solve for x, we can rewrite this as:(3x = 2^4 = 16), so (x = 16/3).But wait, before I conclude, I need to check if this solution is valid in the original equation, because sometimes when manipulating logarithms, we might introduce extraneous solutions or have domain issues. Let's verify.First, check the domain. The original equation has logarithms: (log_{4} x) and (log_{3} x). For (log_{4} x) to be defined, x > 0. For (log_{3} x), x > 0 as well. Also, the exponent on the right side involves (log_{3} x), so x ≠ 1 because the denominator in the exponent would be zero if x=1. But our solution is x=16/3, which is greater than 0 and not 1, so domain-wise it's okay.Now, substitute x = 16/3 into the original equation and check if both sides are equal.Left side: (log_{2} 3 + 2 log_{4} (16/3)).First, compute (log_{4} (16/3)):(log_{4} 16 - log_{4} 3 = 2 - log_{4} 3). Then multiply by 2: 2*(2 - log_{4} 3) = 4 - 2 log_{4} 3.So left side becomes: (log_{2} 3 + 4 - 2 log_{4} 3).Convert (log_{4} 3) to base 2: (log_{4} 3 = frac{log_{2} 3}{2}). So 2 * (log_{4} 3) = (log_{2} 3). Therefore, left side is:(log_{2} 3 + 4 - log_{2} 3 = 4).Right side of the original equation is x^(exponent) = (16/3)^{...} which we previously simplified to 4. So left side is 4, right side is 4. Therefore, it checks out.So x = 16/3 is a valid solution.Wait, but is this the only solution? Let me think. The original equation simplifies to (log_{2}(3x) = 4), which gives x = 16/3 uniquely. So unless there was a mistake in the simplification steps that might have overlooked other solutions, this should be the only one.But let me double-check. Suppose there might be another solution if the exponent and the logarithm somehow cross each other. Let me see. Suppose we didn't simplify the right side and instead tried to solve the equation numerically or by substitution.Alternatively, suppose we let y = log_{3} x, which might lead to a different approach. Let me try that.Let me denote y = log_{3} x, which implies that x = 3^y. Let's substitute this into the original equation.Left side: (log_{2} 3 + 2 log_{4} (3^y)). Let's compute (log_{4} (3^y)):(log_{4} 3^y = y log_{4} 3 = y * frac{log_{2} 3}{log_{2} 4} = y * frac{log_{2} 3}{2}). Therefore, 2 * (log_{4} 3^y) = 2 * (y * (frac{log_{2} 3}{2})) = y log_{2} 3.So the left side becomes: (log_{2} 3 + y log_{2} 3 = log_{2} 3 (1 + y)).The right side: x^{frac{log_{9} 16}{log_{3} x}} = (3^y)^{frac{log_{9}16}{y}}.Simplify the exponent: (frac{log_{9}16}{y}). Let's compute (log_{9}16). As before, 9 is 3^2, 16 is 2^4.(log_{9}16 = frac{log_{3}16}{log_{3}9} = frac{log_{3}2^4}{2} = frac{4 log_{3}2}{2} = 2 log_{3}2).Therefore, the exponent becomes (frac{2 log_{3}2}{y}). So the right side is (3^y)^{(2 log_{3}2)/y} = 3^{y * (2 log_{3}2)/y} = 3^{2 log_{3}2}.Simplify 3^{2 log_{3}2} = (3^{log_{3}2})^2 = 2^2 = 4. So the right side is 4.Therefore, the equation becomes:(log_{2}3 (1 + y) = 4).But since y = log_{3}x, substitute back:(log_{2}3 (1 + log_{3}x) = 4).Let me compute 1 + log_{3}x. Since 1 = log_{3}3, so 1 + log_{3}x = log_{3}3 + log_{3}x = log_{3}(3x). Therefore, the left side is:(log_{2}3 * log_{3}(3x)).Using the change of base formula, (log_{3}(3x) = frac{log_{2}(3x)}{log_{2}3}). Therefore, the left side becomes:(log_{2}3 * frac{log_{2}(3x)}{log_{2}3} = log_{2}(3x)).So we end up with (log_{2}(3x) = 4), which is the same equation as before, leading to 3x = 16, so x = 16/3. Therefore, this substitution method also leads to the same unique solution. So it seems that x = 16/3 is indeed the only solution.Therefore, after careful analysis, the solution is x = 16/3.Final AnswerThe solution to the equation is (boxed{dfrac{16}{3}}).

✅:Given the equation:[log_{2} 3 + 2 log_{4} x = x^{frac{log_{9} 16}{log_{3} x}}]We start by simplifying each term. For the left side, we convert the base 4 logarithm to base 2:[2 log_{4} x = 2 cdot frac{log_{2} x}{2} = log_{2} x]Thus, the left side becomes:[log_{2} 3 + log_{2} x = log_{2} (3x)]Next, we simplify the right side. The exponent is:[frac{log_{9} 16}{log_{3} x}]Converting the numerator (log_{9} 16) to base 3:[log_{9} 16 = frac{log_{3} 16}{2} = frac{4 log_{3} 2}{2} = 2 log_{3} 2]The exponent simplifies to:[frac{2 log_{3} 2}{log_{3} x} = 2 log_{x} 2]Thus, the right side becomes:[x^{2 log_{x} 2} = (x^{log_{x} 2})^2 = 2^2 = 4]Equating both sides, we have:[log_{2} (3x) = 4]Solving for (x):[3x = 2^4 implies 3x = 16 implies x = frac{16}{3}]Verification confirms that (x = frac{16}{3}) satisfies the original equation. Therefore, the solution is:[boxed{dfrac{16}{3}}]

🔑:1. Identify the domain of the problem (ODZ, область допустимых значений): [ 0 < x neq 1 ]2. Simplify the logarithmic expression using logarithm properties: [ log _{2} 3 + 2 log _{4} x = x^{frac{log _{9} 16}{log _{3} x}} ] 3. Express the logarithms in a consistent base: Start by converting the base of the logarithm inside the exponent on the right-hand side: [ log_{9} 16 = frac{log_{3} 16}{log_{3} 9} = frac{4 log_{3} 2}{2 log_{3} 3} = 2 log_{3} 2 text{ (since } log_{3} 9 = 2 log_{3} 3text{)} ] Therefore: [ frac{log _{9} 16}{log _{3} x} = frac{2 log_{3} 2}{log _{3} x} = 2 frac{log_{3} 2}{log _{3} x} = 2 log_{x} 2 ] Simplify: [ x^{2 log_{x} 2} = x^{log_{x} 4} text{ (since } log_{x} 2^2 = log_{x} 4text{)} ]4. Rewrite the original equation: [ log _{2} 3 + 2 log _{4} x = x^{log_{x} 4} ]5. Simplify the left-hand side: [ 2 log _{4} x = log _{4} x^2 = log_{2^{2}} x^2 = frac{log_{2} x^2}{log_{2} (2^{2})} = frac{log _{2} x^2}{2} = log_{2} x ] Therefore: [ log_{2} 3 + log_{2} x = 4 quad text{(because } x^{log_{x} 4} = 4 text{)} ]6. Combine the terms on the left-hand side: [ log_{2} (3x) = 4 ]7. Solve for x: [ 3x = 2^4 = 16 ] [ x = frac{16}{3} ]8. Verify the solution in the domain: [ 0 < x neq 1 Rightarrow 0 < frac{16}{3} neq 1 text{, which is true.} ]Conclusion:[boxed{frac{16}{3}}]

🧠:Okay, so I need to prove that if I have positive numbers a₁, a₂, ..., aₙ and their product is 1, then the product (1 + a₁)(1 + a₂)...(1 + aₙ) is at least 2ⁿ. Hmm, let's start by understanding what the problem is asking. First, all the a_i's are positive and multiply to 1. The inequality we need to show is that the product of each (1 + a_i) is greater than or equal to 2ⁿ. Since 2ⁿ is the product of 2 multiplied by itself n times, maybe there's a way to relate each term (1 + a_i) to 2, and then use some inequality to show that when you multiply them all together, you get at least 2ⁿ. I remember that the Arithmetic Mean-Geometric Mean (AM-GM) inequality is often useful in these types of problems. The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. Maybe I can apply AM-GM to each term (1 + a_i). Let's try that. For each individual term 1 + a_i, the arithmetic mean of 1 and a_i is (1 + a_i)/2, and the geometric mean is √(1 * a_i) = √a_i. By AM-GM, (1 + a_i)/2 ≥ √a_i, so multiplying both sides by 2 gives 1 + a_i ≥ 2√a_i. So, if I apply this to each term in the product, I get:(1 + a₁)(1 + a₂)...(1 + aₙ) ≥ (2√a₁)(2√a₂)...(2√aₙ) = 2ⁿ * √(a₁a₂...aₙ)But wait, since a₁a₂...aₙ = 1 (given), then √(a₁a₂...aₙ) = √1 = 1. Therefore, the right-hand side becomes 2ⁿ * 1 = 2ⁿ. So, putting it all together:(1 + a₁)(1 + a₂)...(1 + aₙ) ≥ 2ⁿWhich is exactly what we needed to prove. That seems straightforward. But let me check if there are any conditions or cases where equality holds. Equality in AM-GM occurs when all the terms are equal. So, for each 1 + a_i ≥ 2√a_i, equality holds when 1 = a_i. Therefore, each a_i must be 1. Indeed, if all a_i = 1, then the product (1 + 1)^n = 2ⁿ, which matches the lower bound. So, equality holds when all a_i are 1. But let me think if there could be another scenario where equality holds. Suppose some a_i are not 1 but the product still gives equality. But no, because for each term, unless a_i = 1, the inequality 1 + a_i ≥ 2√a_i is strict. For example, if a_i = 2, then 1 + 2 = 3, and 2√2 ≈ 2.828, so 3 > 2.828. Similarly, if a_i = 1/2, then 1 + 1/2 = 1.5, and 2√(1/2) ≈ 1.414, so 1.5 > 1.414. Therefore, unless all a_i = 1, the inequality is strict. Therefore, equality holds only when all a_i are 1. Is there another way to approach this problem? Maybe induction? Let's try that as a verification. Base case: n = 1. Then we have a₁ = 1 (since the product is 1), and (1 + a₁) = 2, which is equal to 2¹. So the base case holds. Assume that for some k ≥ 1, if a₁a₂...a_k = 1, then (1 + a₁)...(1 + a_k) ≥ 2^k. Now consider the case for k + 1. Let a₁a₂...a_{k+1} = 1. We need to show that (1 + a₁)...(1 + a_{k+1}) ≥ 2^{k+1}. But here, how can we apply the induction hypothesis? Maybe we can group terms. For example, let’s consider a₁a₂...a_k = 1/a_{k+1}. Then, if we set b = a_{k+1}, then a₁a₂...a_k = 1/b. However, this might not directly fit into the induction hypothesis since the product of the first k terms is 1/b, which is not necessarily 1. Unless we can normalize it. Alternatively, maybe use the fact that for any two positive numbers x and y with xy = c, the minimum of (1 + x)(1 + y) occurs when x = y. Wait, that's a different approach. Let me think. Suppose we have two variables x and y such that xy = c. Then, (1 + x)(1 + y) = 1 + x + y + xy = 1 + x + y + c. To minimize this expression given xy = c, we can set x = y = √c by AM-GM, which gives the minimum value. Therefore, (1 + √c)^2. So, in that case, for two variables, the minimal value of the product (1 + x)(1 + y) given xy = c is (1 + √c)^2. But how does this help with induction? Let's try. Suppose for n = k, the statement holds. For n = k + 1, given a₁a₂...a_{k+1} = 1, we can pair two variables together. Let's take a₁ and a₂. If we fix the product a₁a₂ = c, then the minimal value of (1 + a₁)(1 + a₂) is (1 + √c)^2. Then the remaining variables a₃...a_{k+1} have product 1/c. Then, by induction hypothesis, the product (1 + a₃)...(1 + a_{k+1}) ≥ 2^{k - 1} * (1 + √c)^2 / 2²? Wait, maybe not straightforward. Alternatively, maybe if we consider the product (1 + a₁)(1 + a₂)...(1 + a_{k+1}) ≥ 2^{k+1}. Let's use the induction step. Suppose we have k + 1 variables with product 1. Let me set a_{k+1} = 1/(a₁a₂...a_k). Then, the product becomes (1 + a₁)(1 + a₂)...(1 + a_k)(1 + 1/(a₁a₂...a_k)). By the induction hypothesis, the product (1 + a₁)...(1 + a_k) ≥ 2^k, but we need to handle the last term. Hmm, maybe this approach isn't directly helpful. Alternatively, maybe use the inequality we derived earlier via AM-GM. Since that approach worked for any n, and induction might complicate things. But let's see if induction can be made to work. Alternatively, use logarithms. Taking the natural logarithm of both sides:ln[(1 + a₁)(1 + a₂)...(1 + aₙ)] = Σ ln(1 + a_i)We need to show that this sum is ≥ n ln 2. Given that the product of a_i is 1. Maybe use Jensen's inequality? The function f(x) = ln(1 + e^x) is convex? Wait, if we set x_i = ln a_i, then the constraint becomes Σ x_i = 0 (since the product a₁...aₙ = 1 implies the sum of ln a_i = 0). Then, the sum Σ ln(1 + e^{x_i}) ≥ n ln 2. Is f(x) = ln(1 + e^x) convex? Let's check the second derivative. First derivative f’(x) = e^x / (1 + e^x). Second derivative f''(x) = e^x / (1 + e^x)^2, which is positive. Therefore, f(x) is convex. Then, by Jensen's inequality, the average of f(x_i) is ≥ f(average of x_i). Since the average of x_i is (1/n) Σ x_i = 0. Therefore, (1/n) Σ ln(1 + e^{x_i}) ≥ ln(1 + e^{0}) = ln 2. Multiplying both sides by n gives Σ ln(1 + e^{x_i}) ≥ n ln 2, which is exactly what we needed. Therefore, exponentiating both sides gives the original inequality. So, that's another way to prove it using Jensen's inequality. But I think the AM-GM approach is simpler. However, both methods lead to the same conclusion. Wait, but in the Jensen's approach, we have that f(x) = ln(1 + e^x) is convex, so the inequality goes in the right direction. Therefore, since the function is convex, the average of f(x_i) is at least f(average x_i). But since average x_i is 0, then average f(x_i) ≥ ln 2, so sum f(x_i) ≥ n ln 2. This is a valid approach as well. So, either AM-GM or Jensen's inequality can be used. Alternatively, if we consider the substitution where each a_i = x_i / x_{i+1}, but that might complicate things. Wait, no, since the product of all a_i is 1, perhaps we can set variables in a way that a_i = x_{i}/x_{i+1} for i = 1 to n-1, and a_n = x_n/x_1. Then, the product would telescope to 1. However, this substitution might not necessarily help here. Alternatively, consider homogenization. Since the product of a_i is 1, we can write the inequality in a homogeneous form. The left-hand side is the product of (1 + a_i), and the right-hand side is 2ⁿ. Since the product of a_i is 1, perhaps scaling variables. But maybe that's overcomplicating. Another thought: For each term (1 + a_i), since a_i > 0, maybe consider the function f(a_i) = ln(1 + a_i) and analyze its properties. But we already did something similar with Jensen's. Alternatively, consider expanding the product (1 + a₁)(1 + a₂)...(1 + aₙ). The expansion would be the sum over all subsets of the product of a_i's in the subset. For example, for n=2, it's 1 + a₁ + a₂ + a₁a₂. For general n, it's 1 + Σa_i + Σa_ia_j + ... + a₁a₂...aₙ. Since the product a₁a₂...aₙ = 1, the last term is 1. So the expansion is 1 + Σa_i + Σa_ia_j + ... + 1. But how does this help? Maybe each term in the expansion is at least something. For example, by AM-GM, Σa_i ≥ n*(a₁a₂...aₙ)^{1/n} = n*1 = n. Similarly, Σa_ia_j ≥ C(n,2)*(a₁a₂...aₙ)^{2/n} = C(n,2)*1^{2/n} = C(n,2). Wait, but 1^{2/n} is 1, so for each term, the sum of products of k terms would be ≥ C(n,k). Then, the entire expansion would be Σ_{k=0}^n C(n,k) = 2ⁿ. Hence, the product (1 + a₁)...(1 + aₙ) ≥ 2ⁿ. Wait, this seems like a possible approach. Let's check it. Indeed, by the AM-GM inequality, for each subset S of size k, the product of the a_i's in S is at least 1^{k/n} = 1, but wait, actually, no. Wait, if we take all the C(n,k) terms in the sum Σa_{i1}a_{i2}...a_{ik}}, then by AM-GM, each term a_{i1}a_{i2}...a_{ik}} ≥ 1^{k/n}? No, that's not correct. Because the geometric mean of the products would be (Πa_{i1}a_{i2}...a_{ik}})^{1/C(n,k)}. Hmm, this seems complicated. Wait, but maybe each term a_{i1}a_{i2}...a_{ik}} ≥ 1^{k/n} by AM-GM. Wait, but if we consider that the product of all a_i is 1, then for any subset S of size k, the product Π_{i in S} a_i = 1 / Π_{i not in S} a_i. But I don't see how that helps. Alternatively, maybe if we fix k, then the sum Σ_{|S|=k} Π_{i in S} a_i ≥ C(n,k). This is similar to the inequality that the sum of the products over all k-element subsets is at least C(n,k), given that the product of all a_i is 1. Is that a known inequality? Let me think. If we use the AM-GM inequality on the sum of products. For each k, the sum of the products over all k-element subsets is ≥ C(n,k)*(a₁a₂...aₙ)^{C(n-1,k-1)/C(n,k)}}. Wait, that exponent seems complicated. Alternatively, maybe use the inequality that for symmetric sums. Alternatively, since the product of all a_i is 1, perhaps by Maclaurin's inequality. Maclaurin's inequality states that for symmetric sums,S_k / C(n,k) ≥ (S_{k-1}/C(n,k-1)))^{(k)/(k-1)}}But this might not directly apply here. Alternatively, consider that each symmetric sum S_k = Σ_{|S|=k} Π_{i in S} a_i. Then, Maclaurin's inequality says that (S_k / C(n,k))^{1/k} ≥ (S_{k+1}/C(n,k+1))^{1/(k+1)}}But this also might not directly give the desired result. Alternatively, if all a_i = 1, then each S_k = C(n,k). If the a_i's deviate from 1, some terms in S_k increase and others decrease. But given that the product is fixed at 1, by the AM-GM inequality, the symmetric sums are minimized when all variables are equal. Therefore, the minimal value of each S_k is C(n,k), achieved when all a_i = 1. Therefore, S_k ≥ C(n,k) for each k. If this is true, then the expansion of (1 + a₁)...(1 + aₙ) = Σ_{k=0}^n S_k ≥ Σ_{k=0}^n C(n,k) = 2ⁿ. Hence, the inequality holds. But is it true that S_k ≥ C(n,k) for each k, given that the product of a_i's is 1? Let's check with n=2. If a₁a₂=1, then S₁ = a₁ + a₂ ≥ 2 by AM-GM. S₂ = a₁a₂ =1. So Σ_{k=0}^2 S_k =1 + (a₁ + a₂) +1 ≥1 +2 +1=4=2². Which matches. For n=3, if a₁a₂a₃=1, then S₁ = a₁ + a₂ + a₃ ≥3 by AM-GM. S₂ = a₁a₂ +a₁a₃ +a₂a₃ ≥3*(a₁a₂a₃)^{2/3}=3*1=3. S₃ =1. So total is1 +3 +3 +1=8=2³. So seems to work. So, in general, if for each k, S_k ≥ C(n,k), then the total product is ≥2ⁿ. But how to prove that S_k ≥ C(n,k) for each k?This seems like a generalization of AM-GM. For the case k=1, S₁ = Σa_i ≥ n by AM-GM. For k=2, we can use the inequality that the sum of products Σa_ia_j ≥ C(n,2) when the product of all a_i is 1. Wait, but how?Wait, let's take logarithms. Let’s denote the variables x_i = ln a_i. Then, the condition becomes Σx_i =0. The sum S_k is the sum of exp(Σ_{i in S}x_i) over all subsets S of size k. We need to show that this sum is at least C(n,k). But this is equivalent to saying that the average of exp(Σ_{i in S}x_i) over all subsets S of size k is at least 1. Because C(n,k) * average = sum ≥ C(n,k), so average ≥1. But by Jensen's inequality, since exp is convex, the average of exp(Σx_i over S) is ≥ exp(average of Σx_i over S). The average of Σx_i over all subsets S of size k is equal to (k/n) Σx_i = (k/n)*0 =0. Therefore, exp(0)=1. Hence, the average is ≥1, so the sum is ≥C(n,k). Therefore, this shows that S_k ≥ C(n,k) for each k. Hence, when we expand (1 + a₁)...(1 + aₙ) = Σ_{k=0}^n S_k ≥ Σ_{k=0}^n C(n,k) =2ⁿ. This is a more general approach using Jensen's inequality on the exponential function, applied to each symmetric sum. Therefore, multiple approaches confirm the inequality: AM-GM applied directly to each term, induction (though that got a bit stuck), Jensen's inequality on the logarithm function, and using symmetric sums with Jensen's inequality. So the most straightforward proof is using AM-GM on each (1 + a_i) term, multiplying them together, and using the given condition that the product of a_i's is 1. This gives the desired inequality directly. I think this covers the proof thoroughly. The key insight is recognizing that each term (1 + a_i) can be bounded below by 2√a_i via AM-GM, and since the product of all √a_i is 1, the inequality simplifies to 2ⁿ. Equality holds when all a_i =1, which makes sense as that's when each term (1 + a_i) equals 2.

✅:To prove that for positive numbers (a_1, a_2, ldots, a_n) with (a_1a_2 ldots a_n = 1), the inequality ((1+a_1)(1+a_2) ldots (1+a_n) geq 2^n) holds, we can use the Arithmetic Mean-Geometric Mean (AM-GM) Inequality as follows:1. Apply AM-GM to Each Term: For each (i), consider the numbers (1) and (a_i). By AM-GM: [ frac{1 + a_i}{2} geq sqrt{1 cdot a_i} implies 1 + a_i geq 2sqrt{a_i}. ] 2. Multiply All Inequalities: Multiply the inequalities for (i = 1, 2, ldots, n): [ prod_{i=1}^n (1 + a_i) geq prod_{i=1}^n 2sqrt{a_i} = 2^n sqrt{a_1a_2 ldots a_n}. ] 3. Simplify Using Given Condition: Since (a_1a_2 ldots a_n = 1), substitute into the inequality: [ 2^n sqrt{1} = 2^n. ] 4. Combine Results: Thus: [ (1+a_1)(1+a_2) ldots (1+a_n) geq 2^n. ] Equality Condition: Equality occurs if and only if (a_i = 1) for all (i), ensuring (1 + a_i = 2) for each term.Final Answer:[boxed{(1+a_{1})(1+a_{2}) ldots (1+a_{n}) geqslant 2^{n}}]

🔑:To prove that if the numbers ( a_1, ldots, a_n ) are positive and ( a_1 cdots a_n = 1 ), then[(1+a_1)(1+a_2) cdots (1+a_n) geq 2^n,]we start with the given condition ( a_1 a_2 cdots a_n = 1 ).1. Express the Original Product in Terms of Its Reciprocals: Since ( a_1 a_2 cdots a_n = 1 ), we can rewrite the product ((1+a_1)(1+a_2) cdots (1+a_n)) as follows: [ left(1+a_1right)left(1+a_2right) cdots left(1+a_nright) ] Dividing both sides by ( a_1 a_2 cdots a_n ): [ left(1 + a_1right) left(1 + a_2right) cdots left(1 + a_nright) = frac{left(1 + a_1right)left(1 + a_2right) cdots left(1 + a_nright)}{a_1 a_2 cdots a_n} = left(1 + frac{1}{a_1}right) left(1 + frac{1}{a_2}right) cdots left(1 + frac{1}{a_n}right) ]2. Introduce Each Pair Product: For each term (1 + a_i), consider the corresponding term (1 + frac{1}{a_i}). Each pair product can be examined as follows: [ left(1 + a_iright)left(1 + frac{1}{a_i}right) = 1 + a_i + frac{1}{a_i} + 1 = 2 + a_i + frac{1}{a_i} ]3. Applying the Arithmetic Mean-Geometric Mean Inequality: By the AM-GM inequality, we know that: [ a_i + frac{1}{a_i} geq 2 ] Therefore, [ 2 + a_i + frac{1}{a_i} geq 2 + 2 = 4 ] Hence, [ left(1 + a_iright)left(1 + frac{1}{a_i}right) geq 4 ]4. Formulate the Entire Expression Using Pair Products: Use the above results for each (i = 1, 2, ldots, n): [ left((1 + a_1)(1 + frac{1}{a_1})right) cdots left((1 + a_n)(1 + frac{1}{a_n})right) geq 4^n ]5. Simplify and Finalize the Inequality: We have [ left(left(1+a_1right)left(1+a_2right) cdots left(1+a_nright)right)^2 = left(1 + a_1right)left(1 + frac{1}{a_1}right) cdots left(1 + a_nright)left(1 + frac{1}{a_n}right) geq 4^n ] Finally, taking the square root of both sides: [ left(1+a_1right)left(1+a_2right) cdots left(1+a_nright) geq sqrt{4^n} = 2^n ] Conclusion:[boxed{(1+a_1)(1+a_2) cdots (1+a_n) geq 2^n}]