<s>[INST] Minimum value of $\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}$ 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
$$A\geq \frac{3}{\sqrt[3]{xyz}}=\frac{3}{\sqrt[3]{x+y+z}}.$$
 [/INST] 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!
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