3. $x, y, z$ are positive real numbers, and satisfy $x^{4}+y^{4}+z^{4}=1$, find the minimum value of $\frac{x^{3}}{1-x^{8}}+\frac{y^{3}}{1-y^{8}}+\frac{z^{3}}{1-z^{8}}$. (2000 Jiangsu Province Mathematical Olympiad Problem)
3. Applying the mean value theorem, we get
$-u^{8}\left(1-u^{8}\right)^{8}=\frac{1}{8} \cdot 8 u^{8}\left(1-u^{8}\right)\left(1-u^{8}\right) \cdots\left(1-u^{8}\right) \leqslant$
$$\frac{1}{8} \cdot\left(\frac{8 u^{8}+\left(1-u^{8}\right)+\cdots+\left(1-u^{8}\right)}{9}\right)^{8}=\frac{1}{8}\left(\frac{8}{9}\right)^{9}$$

Therefore,
$$\begin{array}{l}
u\left(1-u^{8}\right) \leqslant \frac{8}{\sqrt[4]{3^{9}}} \\
\frac{x^{-3}}{1-x^{8}}+\frac{y^{3}}{1-y^{8}}+\frac{z^{3}}{1-z^{8}} \geqslant \frac{x^{4}+y^{4}+z^{4}}{8} \cdot \sqrt[4]{3^{4}}=\frac{9}{8} \cdot \sqrt[4]{3}
\end{array}$$

When $x=y=z=\frac{1}{\sqrt[4]{3}}$, equality holds in equation 1, thus, the minimum value sought is $\frac{9}{8} \cdot \sqrt[4]{3}$.