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$x$, $y$ 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.

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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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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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