Problem 1492

Obviously, equation (1)
$$\begin{array}{l} \Leftrightarrow \frac{x^{5}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}}{z^{5}+x^{2}+y^{2}} \\ \geqslant \frac{x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{2}}{z^{5}+x^{2}+y^{2}} \end{array}$$

Lemma: For positive real numbers $x, y, z$ satisfying $xyz \geqslant 1$, and real numbers $a, b$ satisfying $a > b \geqslant 0$, prove:
$$x^{a}+y^{a}+z^{a} \geqslant x^{b}+y^{b}+z^{b}.$$

Proof of the lemma: Without loss of generality, assume $x \geqslant y \geqslant z$, then
$$x^{b} \geqslant y^{b} \geqslant z^{b}, \quad x^{a-b} \geqslant y^{a-b} \geqslant z^{a-b}.$$

By the rearrangement inequality, we have
\begin{array}{l} x^{b} y^{a-b} + y^{b} z^{a-b} + z^{b} x^{a-b} \leqslant x^{a} + y^{a} + z^{a}, \\ x^{b} z^{a-b} + y^{b} x^{a-b} + z^{b} y^{a-b} \leqslant x^{a} + y^{a} + z^{a}. \\ \text{Thus, }\left(x^{b} + y^{b} + z^{b}\right)\left(x^{a-b} + y^{a-b} + z^{a-b}\right) \\ = \left(x^{a} + y^{a} + z^{b}\right) + \left(x^{b} y^{a-b} + y^{b} z^{a-b} + \\ \left. z^{b} x^{a-b}\right) + \left(x^{b} z^{a-b} + y^{b} x^{a-b} + z^{b} y^{a-b}\right) \\ \leqslant 3\left(x^{a} + y^{a} + z^{a}\right). \\ \text{Also, }\left(x^{b} + y^{b} + z^{b}\right)\left(x^{a-b} + y^{a-b} + z^{a-b}\right) \\ \geqslant \left(x^{b} + y^{b} + z^{b}\right) 3 \sqrt[3]{x^{a-b} y^{a-b} z^{a-b}} \\ \geqslant 3\left(x^{b} + y^{b} + z^{b}\right), \\ x^{a} + y^{a} + z^{a} \geqslant x^{b} + y^{b} + z^{b}. \end{array}