Example 5 Given $x, y, z \in R^{+}, x^{2}+y^{2}+z^{2}=$ 1 , find the minimum value of $\frac{x^{5}}{y^{2}+z^{2}-y z}+\frac{y^{5}}{x^{2}+y^{2}-x y}+\frac{z^{5}}{x^{2}+y^{2}-x y}$. (2005 Jiangsu Province Mathematical Olympiad Winter Camp)
Solution: By the power mean inequality and a variant of Schur's inequality
(1) we get $\frac{x^{5}}{y^{2}+z^{2}-y z}+\frac{y^{5}}{x^{2}+y^{2}-x y}+$ $\frac{z^{5}}{x^{2}+y^{2}-x y}=\frac{x^{6}}{x y^{2}+x z^{2}-x y z}+$ $\frac{y^{6}}{y x^{2}+y z^{2}-x y z}+\frac{z^{6}}{z x^{2}+z y^{2}-x y z} \geqslant$ $\frac{\left(x^{3}+y^{3}+z^{3}\right)^{2}}{x y^{2}+y x^{2}+y^{2} z+y z^{2}+x z^{2}+z x^{2}-3 x y z} \geqslant$
$\frac{\left(x^{3}+y^{3}+z^{3}\right)^{2}}{x^{3}+y^{3}+z^{3}}=\left(x^{2}\right)^{\frac{3}{2}}+\left(y^{2}\right)^{\frac{3}{2}}+\left(z^{2}\right)^{\frac{3}{2}} \geqslant 3$ $\left(\frac{x^{2}+y^{2}+z^{2}}{3}\right)^{\frac{3}{2}}=3^{-\frac{1}{2}}=\frac{\sqrt{3}}{3}$. Therefore, the minimum value of $\frac{x^{5}}{y^{2}+z^{2}-y z}$ $+\frac{y^{5}}{x^{2}+y^{2}-x y}+\frac{z^{5}}{x^{2}+y^{2}-x y}$ is $\frac{\sqrt{3}}{3}$.