6. Let $x, y, z \in (0,1)$, satisfy
$$
\sqrt{\frac{1-x}{y z}}+\sqrt{\frac{1-y}{z x}}+\sqrt{\frac{1-z}{x y}}=2 \text {. }
$$

Find the maximum value of $x y z$.
(Tang Lihua, provided)
6. Let $u=\sqrt[5]{x y z}$. Then, by the condition and the AM-GM inequality, we have
$$
\begin{array}{l}
2 u^{3}=2 \sqrt{x y z}=\frac{1}{\sqrt{3}} \sum \sqrt{x(3-3 x)} \\
\leqslant \frac{1}{\sqrt{3}} \sum \frac{x+(3-3 x)}{2}=\frac{3 \sqrt{3}}{2}-\frac{1}{\sqrt{3}}(x+y+z) \\
\leqslant \frac{3 \sqrt{3}}{2}-\sqrt{3} \cdot \sqrt[3]{x y z}=\frac{3 \sqrt{3}}{2}-\sqrt{3} u^{2} .
\end{array}
$$

Thus, $4 u^{3}+2 \sqrt{3} u^{2}-3 \sqrt{3} \leqslant 0$, which means
$$
(2 u-\sqrt{3})\left(2 u^{2}+2 \sqrt{3} u+3\right) \leqslant 0 \text {. }
$$

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