2. Let $x, y, z$ be positive numbers, and $x^{2}+y^{2}+z^{2}=1$, find the minimum value of $\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}$. (30th IMO Canadian Training Problem)
2. When $x=y=z=\frac{\sqrt{3}}{3}$, $\frac{x}{1-x^{2}}=\frac{y}{1-y^{2}}=\frac{z}{1-z^{2}}=\frac{\sqrt{3}}{2}$, we conjecture that the minimum value is $\frac{3 \sqrt{3}}{2}$. Below, we prove that if $x, y, z$ are positive numbers, and $x^{2}+y^{2}+z^{2}=1$, then
$$\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}} \geqslant \frac{3 \sqrt{3}}{2}$$

By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
{\left[x^{3}\left(1-x^{2}\right)+y^{3}\left(1-y^{2}\right)+z^{3}\left(1-z^{2}\right)\right]\left(\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}\right) \geqslant} \\
\left(x^{2}+y^{2}+z^{2}\right)^{2}=1
\end{array}$$

The expression $x^{3}\left(1-x^{2}\right)+y^{3}\left(1-y^{2}\right)+z^{3}\left(1-z^{2}\right)$ is derived from the idea of using the Cauchy-Schwarz inequality based on $x^{2}+y^{2}+z^{2}=1$.

If we can prove
$$x^{3}\left(1-x^{2}\right)+y^{3}\left(1-y^{2}\right)+z^{3}\left(1-z^{2}\right) \leqslant \frac{2}{3 \sqrt{3}}$$

then our conjecture is valid. Note that (1) is equivalent to
$$x^{5}+y^{5}+z^{5}+\frac{2}{3 \sqrt{3}} \geqslant x^{3}+y^{3}+z^{3}$$

Using the condition $x^{2}+y^{2}+z^{2}=1$, the above inequality becomes
$$\left(x^{5}+\frac{2}{3 \sqrt{3}} x^{2}\right)+\left(y^{5}+\frac{2}{3 \sqrt{3}} y^{2}\right)+\left(z^{5}+\frac{2}{3 \sqrt{3}} z^{2}\right) \geqslant x^{3}+y^{3}+z^{3}$$

By the AM-GM inequality, we have
$$x^{5}+\frac{2}{3 \sqrt{3}} x^{2}=x^{5}+\frac{1}{3 \sqrt{3}} x^{2}+\frac{1}{3 \sqrt{3}} x^{2} \geqslant 3 \sqrt[3]{x^{5} \cdot \frac{1}{3 \sqrt{3}} x^{2} \cdot \frac{1}{3 \sqrt{3}} x^{2}}=x^{3}$$

Similarly, $y^{5}+\frac{2}{3 \sqrt{3}} y^{2} \geqslant y^{3}, z^{5}+\frac{2}{3 \sqrt{3}} z^{2} \geqslant z^{3}$, so (2) holds, and thus (1) holds.
In conclusion, the minimum value is $\frac{3 \sqrt{3}}{2}$, and it is achieved if and only if $x=y=z=\frac{\sqrt{3}}{3}$.