For positive numbers $x$, $y$, and $z$ satisfying $x^{2}+y^{2}+z^{2}=1$, find
$$\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}$$

the minimum value.
Solution: Clearly, $x \cdot y, z \in (0,1)$,
By the AM-GM inequality for three variables,
$2 x^{2}\left(1-x^{2}\right)\left(1-x^{2}\right) \leqslant\left(\frac{2}{3}\right)^{3}=\frac{8}{27}$,
with equality if and only if $2 x^{2}=1-x^{2}$, i.e., $x^{2}=\frac{1}{3}$.
$$\begin{array}{l}
2 x^{2}\left(1-x^{2}\right)\left(1-x^{2}\right) \leqslant \frac{8}{27} \Leftrightarrow x\left(1-x^{2}\right) \leqslant \frac{2}{3 \sqrt{3}} \\
\quad \Leftrightarrow \frac{x}{1-x^{2}} \geqslant \frac{3 \sqrt{3}}{2} x^{2},
\end{array}$$

Similarly, $\frac{y}{1-y^{2}} \geqslant \frac{3 \sqrt{3}}{2} y^{2}$,
(F)
$$\frac{z}{1-z^{2}} \geqslant \frac{3 \sqrt{3}}{2} z^{2}$$
Adding (2), (3), and (1) yields
$$\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}} \geqslant \frac{3 \sqrt{3}}{2}.$$

Equality holds if and only if $x^{2}=y^{2}=z^{2}=\frac{1}{3}$. Therefore,
$$\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}$$

has a minimum value of $\frac{3}{2} \frac{\sqrt{3}}{2}$.