Minimum value of $\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}$

Question

Let $x,y,z>0$ and $x+y+z=xyz$. What is the minimum value of $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}?$$

In the case when $x=y=z$, the equation $x+y+z=xyz$ translates to $3x=x^3$, or $x=\sqrt{3}$. If $A$ denotes the quantity that we want to minimize, then $A=\sqrt{3}$ as well.

If we use the inequality of arithmetic and geometric means, we get $$A\geq \frac{3}{\sqrt[3]{xyz}}=\frac{3}{\sqrt[3]{x+y+z}}.$$

Answer 1

The constraint allows the substitution $x=\tan A, y = \tan B, z = \tan C$ for some acute triangle $\triangle ABC$. Further from rearrangement inequality we have: $$\sum_{cyc} \frac{x}{y^2} \geqslant \sum_{cyc} \frac1x = \sum_{cyc} \cot A$$

Now $x \mapsto \cot x$ is convex for $x \in (0, \frac{\pi}2)$, so we may use Jensen to conclude $$\sum_{cyc} \cot A \geqslant 3 \cot \frac{\pi}3$$ Equality is when $x=y=z$.

Answer 2

From our constraint $x + y + z = xyz$, we obtain that $z = \frac{x+y}{xy-1}$, plugging that into $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}$$ we obtain $$\frac{x}{y^2} + \frac{y(xy-1)^2}{(x+y)^2} + \frac{x+y}{x^2 (xy-1)}$$ Call this $f(x,y)$. We can then find a minimum value by doing the second derivative test. First, we need to find the critical points by setting $$f(x,y) = \frac{x}{y^2} + \frac{y(xy-1)^2}{(x+y)^2} + \frac{x+y}{x^2 (xy-1)} = 0$$ We would obtain some ordered pair $(a, b)$ (there may be more than one set of critical points - there may be infinitely many even!). From this we calculate the Hessian, $$H = \det \begin{pmatrix} f_{xx}(a,b) & f_{xy}(a,b) \\ f_{yx}(a,b) & f_{yy}(a,b)\end{pmatrix}$$ For a minimum we want $H > 0$ and $f_{xx}(a,b) > 0$, once we find that culprit, we have found our minimum.

Answer 3

For $x=y=z=\sqrt3$ we get a value $\sqrt3$.

We'll prove that it's a minimal value.

Indeed, we need to prove that $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\geq\sqrt{\frac{3(x+y+z)}{xyz}}$$ or $$\sum_{cyc}\left(\frac{x}{y^2}-\frac{2}{y}+\frac{1}{x}\right)+\sum_{cyc}\frac{1}{x}-\sqrt{\frac{3(x+y+z)}{xyz}}\geq0$$ or $$\sum_{cyc}\frac{(x-y)^2}{y^2x}+\frac{xy+xz+yz-\sqrt{3xyz(x+y+z)}}{xyz}\geq0$$ or $$\sum_{cyc}\frac{(x-y)^2}{y^2x}+\frac{\sum\limits_{cyc}z^2(x-y)^2}{2xyz\left(xy+xz+yz+\sqrt{3xyz(x+y+z)}\right)}\geq0.$$ Done!

Answer 4

Let $x_1 = 1/x, y_1 = 1/y, z_1 = 1/z$. Then $x_1y_1 + x_1z_1 + y_1z_1 = 1$.

$$ \sum_{cyc} \frac{x}{y^2} \geq \sum_{cyc} \frac1x = x_1 + y_1 + z_1 $$

Due to rearrangement inequality $(x_1 + y_1 + z_1)^2 \geq 3(x_1y_1 + x_1z_1 + y_1z_1) = 3$. So

$$ x_1 + y_1 + z_1 \geq \sqrt{3} $$