1.5.11 * Let positive real numbers $x, y, z$ satisfy: $x y z=1$. Find the minimum value of the algebraic expression
$$
\frac{x^{3}}{(1+x)(1+y)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+y)(1+z)}
$$
When $x=y=z=1$, the value of (1) is $\frac{3}{4}$. Note that
$$
\begin{array}{l}
\frac{x^{3}}{(1+x)(1+y)}+\frac{1+x}{8}+\frac{1+y}{8} \geqslant 3 \sqrt[3]{\frac{x^{3}}{(1+x)(1+y)} \cdot \frac{1+x}{8} \cdot \frac{1+y}{8}}=\frac{3}{4} x, \\
\frac{y^{3}}{(1+z)(1+x)}+\frac{1+z}{8}+\frac{1+x}{8} \geqslant \frac{3}{4} y, \frac{z^{3}}{(1+y)(1+z)}+\frac{1+y}{8}+\frac{1+z}{8} \geqslant \frac{3}{4} z .
\end{array}
$$

Summing up the above three inequalities, we know that (1) $\geqslant \frac{1}{2}(x+y+z)-\frac{3}{4}$
$$
\geqslant \frac{1}{2} \cdot 3 \sqrt[3]{x y z}-\frac{3}{4}=\frac{3}{2}-\frac{3}{4}=\frac{3}{4} .
$$

Therefore, the minimum value sought is $\frac{3}{4}$.
Comment: The method of balancing terms used here is very common in the application of the AM-GM inequality.