1. $x, y, z$ are positive real numbers, and satisfy $x^{4}+y^{4}+z^{4}=1$, find the minimum value of $\frac{x^{3}}{1-x^{8}}+\frac{y^{3}}{1-y^{8}}+\frac{z^{3}}{1-z^{8}}$. (1999 Jiangsu Province Mathematical Winter Camp Problem)
1. $\frac{x^{3}}{1-x^{8}}+\frac{y^{3}}{1-y^{8}}+\frac{z^{3}}{1-z^{8}} \geqslant \frac{x^{4}}{x\left(1-x^{8}\right)}+\frac{y^{4}}{y\left(1-y^{8}\right)}+\frac{z^{4}}{z\left(1-z^{8}\right)}$.

Let $f(u)=u\left(1-u^{8}\right), u \in(0,1)$, we find the maximum value of $f(u)$. By $A_{9} \geqslant G_{9}$ we get $8=8 u^{8}+\left(1-u^{8}\right)+\left(1-u^{8}\right)+\cdots+\left(1-u^{8}\right) \geqslant 9 \cdot \sqrt[9]{8 u^{8}\left(1-u^{8}\right)^{8}}$, thus $f(u) \leqslant$ $\frac{8}{\sqrt[4]{3^{9}}}$, which means when $u=\sqrt[4]{\frac{1}{3}}$ the maximum value of $f(u)$ is $\frac{8}{\sqrt[4]{3^{9}}}$. Therefore, $\frac{x^{3}}{1-x^{8}}+\frac{y^{3}}{1-y^{8}}+\frac{z^{3}}{1-z^{8}} \geqslant$ $\frac{\sqrt[4]{3^{9}}}{8}\left(x^{4}+y^{4}+z^{4}\right)=\frac{9}{8} \cdot \sqrt[4]{3}$.

When $x=y=z=\sqrt[4]{\frac{1}{3}}$, the above inequality holds with equality, hence the minimum value of $\frac{x^{3}}{1-x^{8}}+\frac{y^{3}}{1-y^{8}}+\frac{z^{3}}{1-z^{8}}$ is $\frac{9}{8} \cdot \sqrt[4]{3}$.