1.5.9 ** Positive real numbers $x, y, z$ satisfy: $x^{4}+y^{4}+z^{4}=1$. Find the minimum value of the expression $\frac{x^{3}}{1-x^{8}}+\frac{y^{3}}{1-y^{8}}+$ $\frac{z^{3}}{1-z^{8}}$.
Given that $x, y, z \in (0,1)$, using the conclusion from the previous problem, we know that the maximum value of $x(1-x^8)$ is $8 \cdot \left(\frac{1}{9}\right)^{\frac{9}{8}}$. Therefore, $\frac{x^3}{1-x^8} = \frac{x^4}{x(1-x^8)} \geqslant \frac{9^{\frac{9}{8}}}{8} x^4$. Similarly, we can get $\frac{y^3}{1-y^8} \geqslant \frac{9^{\frac{9}{8}}}{8} y^4$, $\frac{z^3}{1-z^8} \geqslant \frac{9^{\frac{9}{8}}}{8} z^4$. Adding these three inequalities, and using $x^4 + y^4 + z^4 = 1$, we have
$$
\frac{x^3}{1-x^8} + \frac{y^3}{1-y^8} + \frac{z^3}{1-z^8} \geqslant \frac{9^{\frac{9}{8}}}{8} = \frac{9}{8} \cdot \sqrt[4]{3},
$$
with equality when $x = y = z = \sqrt[4]{\frac{1}{3}}$. Therefore, the minimum value we are looking for is $\frac{9 \cdot \sqrt[4]{3}}{8}$.