4.3.78 㮡 Positive real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=1$. Find the minimum value of $\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}$.
When $x=y=z=\frac{\sqrt{3}}{3}$, $\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}=\frac{3 \sqrt{3}}{2}$.
Below we prove that for any $x, y, z$ satisfying the given conditions, we have
$$
\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}} \geqslant \frac{3 \sqrt{3}}{2} .
$$

According to the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
& {\left[x^{3}\left(1-x^{2}\right) \mid y^{3}\left(1-y^{2}\right)+z^{3}\left(1-z^{2}\right)\right]\left(\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}\right) } \\
\geqslant & {\left[\sqrt{x^{3}\left(1-x^{2}\right)} \sqrt{\frac{x}{1-x^{2}}}+\sqrt{y^{3}\left(1-y^{2}\right)} \sqrt{\frac{y}{1-y^{2}}}+\sqrt{z^{3}\left(1-z^{2}\right)} \sqrt{\frac{z}{1-z^{2}}}\right]^{2} } \\
= & \left(x^{2}+y^{2}+z^{2}\right)^{2}=1,
\end{aligned}
$$

Therefore, it suffices to prove that $x^{3}\left(1-x^{2}\right)+y^{3}\left(1-y^{2}\right)+z^{3}\left(1-z^{2}\right) \leqslant \frac{2}{3 \sqrt{3}}$, i.e.,
$$
x^{5}+y^{5}+z^{5}+\frac{2}{3 \sqrt{3}} \geqslant x^{3}+y^{3}+z^{3} .
$$

Using the given condition $x^{2}+y^{2}+z^{2}=1$, the above inequality can be rewritten as
$$
\left(x^{5}+\frac{2}{3 \sqrt{3}} x^{2}\right)+\left(y^{5}+\frac{2}{3 \sqrt{3}} y^{2}\right)+\left(z^{5}+\frac{2}{3 \sqrt{3}} z^{2}\right) \geqslant x^{3}+y^{3}+z^{3} .
$$

By the AM-GM inequality,
$$
x^{5}+\frac{2}{3 \sqrt{3}} x^{2}=x^{5}+\frac{1}{3 \sqrt{3}} x^{2}+\frac{1}{3 \sqrt{3}} x^{2} \geqslant 3 \sqrt[3]{\frac{1}{27} x^{9}}=x^{3} .
$$

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 (1) holds, with equality if and only if $x=y=z=\frac{\sqrt{3}}{3}$. Therefore, the minimum value is $\frac{3 \sqrt{3}}{2}$.

Comments:
(1) It is quite challenging to directly find the minimum value through inequalities. We first guess that the minimum value is $\frac{3 \sqrt{3}}{2}$, and then prove an inequality, which is also an effective method for finding the minimum value.
(2) When applying the Cauchy-Schwarz inequality, a crucial step is to grasp the relationship between the degrees of the quantities on both sides of the inequality. In this problem, to utilize the condition $x^{2}+y^{2}+z^{2}=1$, we constructed the expression $x^{3}\left(1-x^{2}\right)+y^{3}\left(1-y^{2}\right)+z^{3}\left(1-z^{2}\right)$.