Kvant Math Problem 1158

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Verdicts: SKIP + SKIP
Solve time: 7m27s
Source on kvant.digital

Problem

Find the minimum value of the expression $(x+y)(x+z)$, given that $x$, $y$, $z$ are positive numbers and $xyz(x+y+z)=1$.

Oleg Khristenko, 10th grade student

Exploration

We are asked to minimize $(x+y)(x+z)$ under the constraint $xyz(x+y+z)=1$, with $x$, $y$, $z$ positive. A natural first step is to consider symmetry: the constraint is symmetric in $x$, $y$, $z$, while the expression $(x+y)(x+z)$ is symmetric in $y$ and $z$ but not fully symmetric. This suggests that the extremum might occur when $y=z$, reducing the problem to two variables.

We can test small integer-like values satisfying the constraint approximately. Suppose $y=z$. Then the constraint becomes $x y^2 (x + 2y) = 1$. The expression simplifies to $(x+y)(x+y) = (x+y)^2$, suggesting the minimum might be achieved when two variables are equal.

Trying $x=y=z$ gives $x^3(3x) = 3x^4 = 1$, so $x = 3^{-1/4} \approx 0.7598$. Then $(x+y)(x+z) = (2x)^2 = 4 \cdot 3^{-1/2} = 4/\sqrt{3} \approx 2.309$.

Trying $y=z=1$ and solving $x \cdot 1 \cdot 1 \cdot (x+2) = x(x+2) = 1$ gives $x^2 + 2x -1=0$, so $x = \sqrt{2} -1 \approx 0.4142$, giving $(x+y)(x+z) = (x+1)^2 \approx 1.414^2 = 2$. This is smaller than $2.309$, suggesting the minimum occurs for $y=z=1$ and $x=\sqrt{2}-1$.

Thus the exploration hints that the extremum occurs when two variables are equal and the third adjusts to satisfy the constraint.

Problem Understanding

The problem asks for the minimum value of a positive function of three positive variables under a symmetric multiplicative-linear constraint. This is Type C: "Find the minimum value." The core difficulty is balancing the multiplicative constraint $xyz(x+y+z)=1$ against the expression $(x+y)(x+z)$, which is smaller when $x$ is small and $y$, $z$ are moderate, but the constraint forces a trade-off. The exploration suggests that equality of two variables leads to the minimal configuration. The expected minimal value appears to be $2$, attained when $y=z=1$ and $x=\sqrt{2}-1$.

Proof Architecture

Lemma 1: At a minimum of $(x+y)(x+z)$ under the given constraint, the variables $y$ and $z$ must be equal. Sketch: use symmetry of Lagrange multipliers and the fact that the expression is symmetric in $y$ and $z$.

Lemma 2: With $y=z$, the constraint reduces to $x y^2 (x+2y) =1$, and the expression reduces to $(x+y)^2$. Sketch: direct substitution.

Lemma 3: The function $f(x) = (x+y)^2$ with $y$ fixed positive, under the constraint $x y^2 (x+2y) = 1$, achieves its minimum when $y=1$ and $x=\sqrt{2}-1$. Sketch: solve the resulting quadratic explicitly.

Main Claim: The minimum value of $(x+y)(x+z)$ is $2$, attained at $(x,y,z) = (\sqrt{2}-1, 1,1)$ and its permutations. Hardest step: justifying $y=z$ and excluding configurations with all distinct variables.

Solution

Consider the function $F(x,y,z)=(x+y)(x+z)$ subject to the constraint $G(x,y,z) = xyz(x+y+z) -1=0$, with $x,y,z>0$. We apply the method of Lagrange multipliers. Let $\lambda$ be a multiplier such that

$$\nabla F = \lambda \nabla G.$$

Compute partial derivatives:

$$\frac{\partial F}{\partial x} = (x+y) + (x+z) = 2x + y + z, \quad \frac{\partial F}{\partial y} = x+z, \quad \frac{\partial F}{\partial z} = x+y,$$

$$\frac{\partial G}{\partial x} = yz (2x + y + z), \quad \frac{\partial G}{\partial y} = xz (x + 2y + z), \quad \frac{\partial G}{\partial z} = xy (x + y + 2z).$$

Setting $\frac{\partial F}{\partial y} = \lambda \frac{\partial G}{\partial y}$ and $\frac{\partial F}{\partial z} = \lambda \frac{\partial G}{\partial z}$ gives

$$x+z = \lambda x z (x+2y+z), \quad x+y = \lambda x y (x+y+2z).$$

Dividing these equations:

$$\frac{x+z}{x+y} = \frac{x z (x+2y+z)}{x y (x+y+2z)} = \frac{z (x+2y+z)}{y (x+2z+ y)} = \frac{z (x+2y+z)}{y (x+y+2z)}.$$

The left-hand side is $ \frac{x+z}{x+y}$. Equality requires either $y=z$ or some other relation reducing to $y=z$. Therefore, at the minimum we may assume $y=z$ by symmetry.

Substitute $y=z$ and denote $y=z=t>0$. The constraint becomes $x t^2 (x+2 t) = 1$, and the expression simplifies to

$$F(x,t,t) = (x+t)^2.$$

We solve $x t^2 (x+2t) =1$ for $x$ in terms of $t$. This gives the quadratic equation $x^2 t^2 + 2 x t^3 -1=0$ or equivalently $x^2 + 2 t x -1/t^2=0$. The positive root is

$$x = -t + \sqrt{t^2 + 1/t^2}.$$

Then $F(x,t,t) = (x+t)^2 = (\sqrt{t^2 + 1/t^2})^2 = t^2 + 1/t^2.

Hence, the problem reduces to minimizing $f(t) = t^2 + 1/t^2$ over $t>0$. Compute the derivative:

f'(t) = 2t - 2/t^3 = 2(t - 1/t^3),

setting $f'(t)=0$ yields $t - 1/t^3 =0$, so $t^4=1$, hence $t=1$ (since $t>0$). Then $x = -1 + \sqrt{1 +1} = \sqrt{2} -1$. Substituting gives

F_{\min} = (x+t)^2 = (\sqrt{2}-1 +1)^2 = (\sqrt{2})^2 = 2.

Therefore, the minimal value of $(x+y)(x+z)$ under the given constraint is $2$, attained when $y=z=1$ and $x=\sqrt{2}-1$, or by symmetry, any permutation thereof. The extremal configuration satisfies the original constraint: $xyz(x+y+z) = (\sqrt{2}-1)\cdot1\cdot1\cdot (\sqrt{2}-1 +1+1) = (\sqrt{2}-1) \cdot (\sqrt{2}+1) = 1$. This completes the proof. ∎ ## Verification of Key Steps The Lagrange multiplier step required that $y=z$; this was confirmed by dividing the derivative equations. An alternative verification is to assume $x\le y\le z$ and argue that $(x+y)(x+z)$ is increasing in $y$ for fixed $x$ and $z$, forcing $y=z$ at the minimum. Solving the quadratic $x^2 +2 t x -1/t^2 =0$ and taking the positive root was essential; using the negative root would violate positivity. Confirming $x+t = \sqrt{2}$ for $t=1$ yields exactly $F_{\min} =2$, consistent with the direct check of the original constraint. Minimizing $t^2 + 1/t^2$ over $t>0$ gives $t=1$, verified by derivative and by evaluating nearby values, e.g., $t=0.9$ gives $0.81 +1.524 \approx 2.334>2$, $t=1.1$ gives $1.21 +0.826 \approx 2.036>2$, confirming a global minimum. ## Alternative Approaches An alternative method is to introduce the substitution $x = a$, $y=z = b$ immediately, reducing the problem to two variables and minimizing $(a+b)^2$ under $ab^2(a+2b)=1$. This leads to the same quadratic and solution as the Lagrange