<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] Also, by AM-GM and Muirhead we obtain:
$$\sum_{cyc}\frac{x^2}{x^4+yz}=\sum_{cyc}\frac{\frac{9x^4y^2z^2}{(x^2+y^2+z^2)^2}}{x^4+\frac{9x^2y^3z^3}{(x^2+y^2+z^2)^2}}=\sum_{cyc}\frac{9x^2y^2z^2}{x^2(x^2+y^2+z^2)^2+9y^3z^3}\leq$$
$$\leq\sum_{cyc}\frac{9x^2y^2z^2}{6x(x^2+y^2+z^2)\sqrt{y^3z^3}}=\frac{3\sum\limits_{cyc}x\sqrt{yz}}{2(x^2+y^2+z^2)}\leq\frac{3}{2}.$$
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