Kvant Math Problem 1404

We are asked to maximize the expression

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Problem

Three numbers $x$, $y$, $z$ satisfy the conditions $x+y+z=0$, $xyz=2$. Find the maximum value of $\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}$.

S. Doichev, R. Kozarev

Exploration

We are asked to maximize the expression

$S = \frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x}$

under the constraints $x+y+z=0$ and $xyz=2$. A first step is to consider symmetry. The constraints suggest a relationship similar to Viète's formulas for a cubic with roots $x$, $y$, $z$. Rewriting the first constraint as $z = -x-y$ allows the expression to be written in two variables:

$S = \frac{x^2}{y} + \frac{y^2}{-(x+y)} + \frac{(-x-y)^2}{x}.$

Simplifying leads to

$S = \frac{x^2}{y} - \frac{y^2}{x+y} + \frac{(x+y)^2}{x} = \frac{x^2}{y} + \frac{(x+y)^2}{x} - \frac{y^2}{x+y}.$

Testing small integers seems impossible because $xyz = 2$ is not an integer cube, so rational examples are more promising. We can attempt symmetry patterns: perhaps $x = y \neq z$ or $x = -y$, etc. Setting $x=y$ gives $z=-2x$ and $x^2 y = x^2 x = x^3$; $xyz= x^2(-2x) = -2x^3=2$, giving $x^3=-1$, so $x=-1$, $y=-1$, $z=2$. Substituting produces

$S = \frac{(-1)^2}{-1} + \frac{(-1)^2}{2} + \frac{2^2}{-1} = -1 + \frac{1}{2} - 4 = -\frac{9}{2}.$

This is negative. Perhaps $x = -y$ gives $z=0$, impossible because $xyz=0\neq 2$. Another natural guess is to try two equal positive numbers and one negative, e.g., $x=y=a$, $z=-2a$; we get $xyz=-2a^3$, which equals 2 if $a=-1$, as above. The other symmetry $x=z$ gives $y=-2x$, then $xyz=-2x^3=2$, same as above. So the maximal value is likely positive, suggesting the extremum occurs when all three numbers are distinct.

We can also attempt a Lagrange multiplier approach. Consider $S$ as a function of $x$, $y$, $z$ with constraints $x+y+z=0$ and $xyz=2$. This leads to a system of three rational equations in three variables. A more symmetric idea is to try substitution $z=-x-y$ and define $S$ in terms of $x$ and $y$ only:

$S(x,y) = \frac{x^2}{y} + \frac{y^2}{-x-y} + \frac{(x+y)^2}{x}.$

This is still rational but can be simplified to

$S(x,y) = \frac{x^3 - y^3 + (x+y)^3}{xy(x+y)} = \frac{x^3 + 3x^2y + 3xy^2 + y^3 - y^3 + (x+y)^3?}{xy(x+y)},$

which requires careful algebra. Testing symmetric patterns such as $x=2$, $y=-1$, $z=-1$ gives $xyz=2$ and $x+y+z=0$. Then $S = 4/(-1) + 1/(-1) + 1/2 = -4 -1 + 1/2 = -9/2$ as before.

Trying $x=1$, $y=-2$, $z=1$, $xyz= -2\neq 2$; $x=2$, $y=1$, $z=-3$, $xyz=-6\neq 2$. A more promising pattern is $x=y$, $z$ negative fractional. Then $x^2 z = 2 \implies z = 2/x^2$, $x+y+z = 2x + 2/x^2=0 \implies x^3 = -1$, giving the previous example.

Hence the previous negative solution is the only solution with repeated numbers. Perhaps the maximal value occurs when numbers are distinct and one is very small to make $S$ large. For example, let $x>0$, $y>0$, $z<0$, then $\frac{z^2}{x} >0$, $\frac{y^2}{z}<0$, $\frac{x^2}{y}>0$. Then $S$ can be positive and perhaps large.

A plausible candidate for the maximum is when $y=z$, $x$ negative. Let $y=z=t>0$, $x= -2t$, then $xyz = -2 t^3 = 2$, $t^3 = -1$, $t=-1$, $x=2$. Then $S = x^2 / y + y^2 / z + z^2 / x = 4/(-1) +1/1 +1/2 = -4 +1 +0.5=-2.5$, still negative.

Testing rational fractions, $x= -1$, $y=2$, $z=-1$, sum $0$, product $2$, $S = 1/2 + 4/(-1) +1/(-1) = 0.5 -4 -1=-4.5$. Still negative. Perhaps the maximum is positive and occurs when one number approaches zero. Let $x\to 0^+$, $y\to +\infty$, $z\to -y$, $xyz \to -0 \cdot y \cdot (-y) =0$, cannot match 2. So the maximum is finite and may occur when two numbers equal.

Trying $x= -1$, $y = ?$, $z = ?$, $x+y+z=0$, $xyz=2$, $y+z=1$, $yz=-2$ gives quadratic $t^2 - t -2=0$, $t=(1\pm \sqrt{1+8})/2 = (1\pm 3)/2$, so $t=2$ or $-1$. Then triples $(-1,2,-1)$ or $(-1,-1,2)$, all previous, $S=-4.5$ or -4.5. Symmetry gives same negatives. So the maximum may be achieved in some other pattern.

A more promising approach is to substitute $x = p$, $y = q$, $z = -p-q$, then $xyz = -pq(p+q)=2$, then $pq(p+q) = -2$, and write $S = x^2 / y + y^2 / z + z^2 / x$ in terms of $p,q$, giving

$S = \frac{p^2}{q} + \frac{q^2}{-(p+q)} + \frac{(p+q)^2}{p} = \frac{p^2}{q} - \frac{q^2}{p+q} + \frac{(p+q)^2}{p}.$

Combine terms:

$\frac{(p+q)^2}{p} - \frac{q^2}{p+q} = \frac{(p+q)^3 - pq^2}{p(p+q)}?$

Better to write as

$S = \frac{p^3 + (p+q)^3 - q^3}{pq(p+q)}$

after checking carefully: $x^2 / y = p^2 / q$, $y^2 / z = q^2 / -(p+q) = - q^2 / (p+q)$, $z^2 / x = (p+q)^2 / p$. Then numerator: $p^2(p+q) - q^2 p + (p+q)^2 q?$, messy.

Another approach is to apply the identity

$\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} = \frac{x^3 z + y^3 x + z^3 y}{xyz} = \frac{x^3 z + y^3 x + z^3 y}{2}.$

Then $S = (x^3 z + y^3 x + z^3 y)/2$. Since $x+y+z=0$, $z=-x-y$, $xyz=2$, then $x^3 z + y^3 x + z^3 y = x^3(-x-y) + y^3 x + (-x-y)^3 y = -x^4 - x^3 y + x y^3 - y (x+y)^3$. Expand $-(x+y)^3 y = -y (x^3 + 3x^2y + 3xy^2 + y^3) = -x^3y - 3x^2 y^2 - 3 x y^3 - y^4$. Sum terms: $-x^4 - x^3y + x y^3 - x^3y - 3