Example 7 Let $x, y, z$ be positive numbers, and $x^{2}+y^{2}+z^{2}=1$, find the minimum value of $\frac{x^{5}}{y^{2}+z^{2}-y z}+\frac{y^{5}}{z^{2}+x^{2}-z x}+$ $\frac{z^{5}}{x^{2}+y^{2}-x y}$.
By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(-\frac{x^{6}}{x y^{2}+x z^{2}-x y z}+\frac{y^{6}}{z^{2} y+x^{2} y-x y z}+\frac{z^{6}}{x^{2} z+y^{2} z-x y z}\right) \\
\left(x y^{2}+x z^{2}+z^{2} y+x^{2} y+x^{2} z+y^{2} z-3 x y z\right) \geqslant \\
\left(x^{3}+y^{3}+z^{3}\right)^{2}
\end{array}$$

By Schur's inequality, we have $x^{3}+y^{3}+z^{3}-\left(x y^{2}+x z^{2}+z^{2} y+x^{2} y+x^{2} z+y^{2} z\right)+3 x y z \geqslant 0$ (or by Example 2 in Chapter 1).

Thus, the above inequality can be transformed into
$$\frac{x^{5}}{y^{2}+z^{2}-y z}+\frac{y^{5}}{z^{2}+x^{2}-z x}+\frac{z^{5}}{x^{2}+y^{2}-x y} \geqslant x^{3}+y^{3}+z^{3}$$

By the Power Mean Inequality, $\sqrt[3]{\frac{x^{3}+y^{3}+z^{3}}{3}} \geqslant \sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}$, we get
$$x^{3}+y^{3}+z^{3} \geqslant \frac{\sqrt{3}}{3}$$

The equality holds if and only if $x=y=z$.
Therefore, when $x=y=z=\frac{\sqrt{3}}{3}$, $\frac{x^{5}}{y^{2}+z^{2}-y z}+\frac{y^{5}}{z^{2}+x^{2}-z x}+\frac{z^{5}}{x^{2}+y^{2}-x y}$ achieves its minimum value $\frac{\sqrt{3}}{3}$.