$$\begin{array}{l} \frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}} \geqslant \frac{x^{5}-x^{2} x y z}{x^{5}+\left(y^{2}+z^{2}\right) x y z} \\ =\frac{x^{4}-x^{2} y z}{x^{4}+\left(y^{2}+z^{2}\right) y z} \\ \text { and } \frac{x^{4}-x^{2} y z}{x^{4}+\left(y^{2}+z^{2}\right) y z}=\frac{2\left(x^{4}-x^{2} y z\right)}{2\left[x^{4}+\left(y^{2}+z^{2}\right) y z\right]} \\ \quad \geqslant \frac{2 x^{4}-x^{2}\left(y^{2}+z^{2}\right)}{2 x^{4}+\left(y^{2}+z^{2}\right)^{2}} \end{array}$$

Therefore, to prove the original inequality, it suffices to prove
$$\sum \frac{2 x^{4}-x^{2}\left(y^{2}+z^{2}\right)}{2 x^{4}+\left(y^{2}+z^{2}\right)^{2}} \geqslant 0$$

Make the substitution $\left(x^{2}, y^{2}, z^{2}\right) \rightarrow(x, y, z)$.
Thus, (1) $\Leftrightarrow \sum \frac{2 x^{2}-x(y+z)}{2 x^{2}+(y+z)^{2}} \geqslant 0$
$$\begin{array}{l} \Leftrightarrow f(x, y, z) \\ = \sum\left\{\left[2 x^{2}-x(y+z)\right]\right. \\ {\left[2 y^{2}+(x+z)^{2}\right] } \\ {\left.\left[2 z^{2}+(x+y)^{2}\right]\right\} } \\ \geqslant 0 \end{array}$$

First, calculate $f(1,0,0), f(1,1,0), f(1,1,1)$, to get $a=2, d=24, n=0$.

Second, calculate $f(0,-1,1), f(0,1, \mathbf{i}) \mathrm{i}$, to get the system of equations
$$\left\{\begin{array}{l} 0=4 \times 2-4 b+4 c-24 \\ 8=2 \times 2-2 b-2 c+24 \end{array}\right.$$

Solving, we get $b=3, c=7$.
Finally, calculate $f(-1,1,1), f(-1,1, \mathrm{i})$, to get the system of equations
$$\begin{array}{l} \left\{\begin{array}{l} 64=80+4 e-8 m, \\ -48=-28-6 e+8 m . \end{array}\right. \\ \text { Solving, we get } e=18, m=11, . \\ \text { Hence } f(x, y, z) \\ =2 g_{6,1}+3 g_{6,2}+7 g_{6,3}+24 g_{6,4}+ \\ 18 g_{6,5}+11 g_{6,6} \\ \geqslant 0 . \end{array}$$

Therefore, the original inequality is proved.