6638 ** Let $x, y, z \in \mathbf{R}^{+}, x^{2}+y^{2}+z^{2}=1$. Find
$$
\frac{x^{5}}{y^{2}+z^{2}-y z}+\frac{y^{5}}{z^{2}+x^{2}-z x}+\frac{z^{5}}{x^{2}+y^{2}-x y}
$$

the minimum value.
By the Cauchy-Schwarz inequality, we have
$$
\sum \frac{x^{6}}{x y^{2}+x z^{2}-x y z} \cdot \sum\left(x y^{2}+x z^{2}-x y z\right) \geqslant\left(\sum x^{3}\right)^{2},
$$

Therefore,
$$
\sum \frac{x^{5}}{y^{2}+z^{2}-y z} \geqslant \frac{\left(\sum x^{3}\right)^{2}}{\sum x\left(y^{2}+z^{2}\right)-3 x y z} .
$$

By Schur's inequality \(\sum x(x-y)(x-z) \geqslant 0\), we get \(\sum x^{3}-\sum x\left(y^{2}+z^{2}\right)+3 x y z \geqslant 0\). Therefore,
\(\sum \frac{x^{5}}{y^{2}+z^{2}-y z} \geqslant \frac{\left(\sum x^{3}\right)^{2}}{\sum x^{3}} \geqslant \sum x^{3} \geqslant 3\left(\sqrt{\frac{\sum x^{2}}{3}}\right)^{3}\) (Power Mean Inequality) \(=\frac{\sqrt{3}}{3}\).
When \(x=y=z=\frac{\sqrt{3}}{3}\), the equality holds.
Thus, the minimum value of \(\sum \frac{x^{5}}{y^{2}+z^{2}-y z}\) is \(\frac{\sqrt{3}}{3}\).