For example, $5 x, y, z$ are positive real numbers, and satisfy $x+y+z=1$. Find the minimum value of $\frac{x^{4}}{y\left(1-y^{2}\right)}+\frac{y^{4}}{z\left(1-z^{2}\right)}+$ $\frac{z^{4}}{x\left(1-x^{2}\right)}$. (Mathematics Bulletin Problem)
Solving: According to symmetry, we estimate that equality holds when $x=y=z$. Applying the AM-GM inequality, we get
$$\begin{array}{l}
\frac{x^{4}}{y(1-y)(1+y)}+\frac{y}{8}+\frac{1-y}{16}+\frac{1+y}{32} \geqslant \\
4 \sqrt[4]{\frac{x^{4}}{y(1-y)(1+y)} \cdot \frac{y}{8} \cdot \frac{1-y}{16} \cdot \frac{1+y}{32}}=\frac{x}{2}
\end{array}$$

Similarly,
$$\begin{array}{l}
\frac{y^{4}}{z(1-z)(1+z)}+\frac{z}{8}+\frac{1-z}{16}+\frac{1+z}{32} \geqslant \frac{y}{2} \\
\frac{z^{4}}{x(1-x)(1+x)}+\frac{x}{8}+\frac{1-x}{16}+\frac{1+x}{32} \geqslant \frac{z}{2}
\end{array}$$

Adding the above three inequalities, we get
$$\begin{array}{l}
\frac{x^{4}}{y\left(1-y^{2}\right)}+\frac{y^{4}}{z\left(1-z^{2}\right)}+\frac{z^{4}}{x\left(1-x^{2}\right)}+\frac{x+y+z}{8}+\frac{3-(x+y+z)}{16}+ \\
\frac{3+(x+y+z)}{32} \geqslant \frac{x+y+z}{2}
\end{array}$$

Since $x+y+z=1$, rearranging gives
$$\frac{x^{4}}{y\left(1-y^{2}\right)}+\frac{y^{4}}{z\left(1-z^{2}\right)}+\frac{z^{4}}{x\left(1-x^{2}\right)} \geqslant \frac{1}{8}$$