Example 4 Let $x, y, z > 0, x^{4} + y^{4} + z^{4} = 1$. Find the minimum value of $\sum \frac{x^{3}}{1 - x^{8}}$.
(2000, Jiangsu Province High School Mathematics Competition)
Analysis: Let $x^{4} = a, y^{4} = b, z^{4} = c$, the expression $\sum \frac{x^{3}}{1 - x^{8}} = \sum \frac{\left(x^{4}\right)^{\frac{3}{4}}}{1 - \left(x^{4}\right)^{2}}$ must be transformed into an expression in terms of $a, b,$ and $c$.
Let $x^{4}=a, y^{4}=b, z^{4}=c$. Then
$a$,
$$\begin{array}{l}
\cdot c>0, a+b+c=1, \\
\sum \frac{x^{3}}{1-x^{8}}=\sum \frac{a^{\frac{3}{4}}}{1-a^{2}} .
\end{array}$$

Let $f(x)=\frac{x^{\frac{3}{4}}}{1-x^{2}}$.
It is easy to know that the tangent line equation of $f(x)$ at $x=\frac{1}{3}$ is
$$g(x)=\frac{9 \sqrt[4]{3}}{8}\left(x-\frac{1}{3}\right)+\frac{3 \sqrt[4]{3}}{8}=\frac{9 \sqrt{3}}{8} x .$$

First, we prove: $\frac{x^{\frac{3}{4}}}{1-x^{2}} \geqslant \frac{9 \sqrt[4]{3}}{8} x(0<x<1)$.
$$\begin{array}{l}
\text { Equation (1) } \Leftrightarrow 8 \geqslant 9 \sqrt[4]{3} x^{\frac{1}{4}}\left(1-x^{2}\right) \\
\Leftrightarrow 9 \sqrt[4]{3} x^{\frac{9}{4}}+8 \geqslant 9 \sqrt[4]{3} x^{\frac{1}{4}} \\
\Leftrightarrow\left(\sqrt[4]{3} x^{\frac{1}{4}}\right)^{9}+\underbrace{1+1+\cdots+1}_{8 \uparrow} \geqslant 9 \sqrt[9]{\left(\sqrt[4]{3} x^{\frac{1}{4}}\right)^{9}},
\end{array}$$

This is the AM-GM inequality, which is obviously true. Therefore, Equation (1) holds.
Thus, $\sum \frac{a^{\frac{3}{4}}}{1-a^{2}} \geqslant \frac{9 \sqrt[4]{3}}{8} \sum a=\frac{9 \sqrt[4]{3}}{8}$.
Therefore, the minimum value of $\sum \frac{x^{3}}{1-x^{8}}$ is $\frac{9 \sqrt[4]{3}}{8}$.