<s>[INST] Calculate the maximum value of $\frac{x^2}{x^4 + yz} + \frac{y^2}{y^4 + zx} + \frac{z^2}{z^4 + xy}$ where $x, y, z > 0$ and $x^2 + y^2 + z^2 = 3xyz$. 
$x$, $y$ and $z$ are positives such that $x^2 + y^2 + z^2 = 3xyz$. Calculate the maximum value of $$\large \frac{x^2}{x^4 + yz} + \frac{y^2}{y^4 + zx} + \frac{z^2}{z^4 + xy}$$

This is (obviously) adapted from a recent competition. There ought to be better solutions that the one I have provided below. So if you could, please post them.
 [/INST] We have that $$\frac{x^2}{x^4 + yz} + \frac{y^2}{y^4 + zx} + \frac{z^2}{z^4 + xy} \le \frac{x^2}{2x^2\sqrt{yz}} + \frac{y^2}{2y^2\sqrt{zx}} + \frac{z^2}{2z^2\sqrt{xy}}$$
$$ = \frac{1}{2}\left(\frac{1}{\sqrt{xy}} + \frac{1}{\sqrt{yz}} + \frac{1}{\sqrt{zx}}\right) = \frac{z\sqrt{xy} + x\sqrt{yz} + y\sqrt{zx}}{2xyz} \le \frac{z^2 + x^2 + y^2}{2xyz} = \frac{3xyz}{2xyz} = \frac{3}{2}$$
The equality sign occurs when $x = y = z = 1$.
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