39. 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)
39. 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 $\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+$ $\frac{z}{1-z^{2}} \geqslant \frac{3 \sqrt{3}}{2}$

Let $a=x^{2}, b=y^{2}, c=z^{2}$, the problem is equivalent to proving under the condition $a+b+c=1$, that $\frac{\sqrt{a}}{1-a}+$ $\frac{\sqrt{b}}{1-b}+\sqrt{c}=\frac{3 \sqrt{3}}{1-c}$

Consider the function $y(x)=\frac{\sqrt{x}}{1-x}$ at the point $\left(\frac{1}{3}, \frac{\sqrt{3}}{2}\right)$. Since $f^{\prime}(x)=$ $\frac{1+x}{2 \sqrt{x}(1-x)^{2}}$, we have $\left.f^{\prime}(x)\right|_{x=\frac{1}{3}}=\frac{3 \sqrt{3}}{2}$.

Thus, the tangent line to the curve $y=\frac{\sqrt{x}}{1+x}$ at the point $\left(\frac{1}{3}, \frac{\sqrt{3}}{4}\right)$ is $y=\frac{3 \sqrt{3}}{2}\left(x-\frac{1}{3}\right)+\frac{\sqrt{3}}{2}=$ $\frac{3 \sqrt{3}}{2} x$

On the interval $(0,1)$,
$$\frac{\sqrt{x}}{1-x}-\frac{3 \sqrt{3}}{2} x=\frac{(\sqrt{3} x+2)(\sqrt{3 x}-1)^{2}}{2(1-x)} \geqslant 0$$

Therefore, when $a, b, c \in(0,1)$, we have
$$\begin{array}{l}
\frac{-\sqrt{a}}{1-a} \geqslant \frac{3 \sqrt{3}}{2} a \\
\frac{\sqrt{b}}{1-b} \geqslant \frac{3 \sqrt{3}}{2} b \\
\frac{\sqrt{c}}{1-c} \geqslant \frac{3 \sqrt{3}}{2}
\end{array}$$

Since $a+b+c=1$, we have $\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c} \geqslant \frac{3 \sqrt{3}}{2}(a+b+c)=\frac{3 \sqrt{3}}{2}$