$\square$ Example 7 Let $x, y, z$ be positive numbers, and $x^{2}+y^{2}+z^{2}=1$, find the minimum value of $\frac{x^{5}}{y^{2}+z^{2}-y z}+$ $\frac{y^{5}}{z^{2}+x^{2}-z x}+\frac{z^{5}}{x^{2}+y^{2}-x y}$. (2006 Turkish Mathematical Olympiad Problem)
By Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\quad\left(\frac{x^{6}}{x y^{2}+x z^{2}-x y z}+\frac{y^{6}}{z^{2} y+x^{2} y-x y z}+\frac{z^{6}}{x^{2} z+y^{2} z-x y z}\right)\left(x y^{2}\right. \\
\left.+x z^{2}+z^{2} y+x^{2} y+x^{2} z+y^{2} z-3 x y z\right) \geqslant\left(x^{3}+y^{3}+z^{3}\right)^{2},
\end{array}$$

By Schur's inequality, we have
$$x^{3}+y^{3}+z^{3}-\left(x y^{2}+x z^{2}+z^{2} y+x^{2} y+x^{2} z+y^{2} z\right)+3 x y z \geqslant 0,$$

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 AM-GM 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 $\frac{\sqrt{3}}{3}$.