Example 25 Let $x, y, z \in(0,1)$, satisfying:
$$\sqrt{\frac{1-x}{y z}}+\sqrt{\frac{1-y}{z x}}+\sqrt{\frac{1-z}{x y}}=2,$$

Find the maximum value of $x y z$.
Let $u=\sqrt[6]{x y z}$, then by the condition and the AM-GM inequality, we have
$$\begin{aligned}
2 u^{3}= & 2 \sqrt{x y z}=\frac{1}{\sqrt{3}} \sum \sqrt{x(3-3 x)} \\
\leqslant & \frac{1}{\sqrt{3}} \sum \frac{x+(3-3 x)}{2} \\
& =\frac{3 \sqrt{3}}{2}-\frac{1}{\sqrt{3}}(x+y+z) \\
& \leqslant \frac{3 \sqrt{3}}{2}-\sqrt{3} \cdot \sqrt[3]{x y z} \\
& =\frac{3 \sqrt{3}}{2}-\sqrt{3} u^{2}
\end{aligned}$$

Thus,
$$4 u^{3}+2 \sqrt{3} u^{2}-3 \sqrt{3} \leqslant 0$$

That is,
$$(2 u-\sqrt{3})\left(2 u^{2}+2 \sqrt{3} u+3\right) \leqslant 0$$

Therefore, $u \leqslant \frac{\sqrt{3}}{2}$. From this, we know that $x y z \leqslant \frac{27}{64}$, with equality when
$$x=y=z=\frac{3}{4}$$

Thus, the maximum value sought is $\frac{27}{64}$.