Problem 1504

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
$$\text{x b y a-b + y b z a-b + z b x a-b x a + y a + z a , x b z a-b + y b x a-b + z b y a-b x a + y a + z a . Thus, (x b + y b + z b ) (x a-b + y a-b + z a-b ) = (x a + y a + z b ) + (x b y a-b + y b z a-b + . z b x a-b ) + (x b z a-b + y b x a-b + z b y a-b ) 3 (x a + y a + z a ). Also, (x b + y b + z b ) (x a-b + y a-b + z a-b ) (x b + y b + z b ) 3 [3] x a-b y a-b z a-b 3 (x b + y b + z b ), x a + y a + z a x b + y b + z b .}$$