38. Let $x, y, z \in(0,1)$, and satisfy $\sqrt{\frac{1-x}{y z}}+\sqrt{\frac{1-y}{z x}}+\sqrt{\frac{1-z}{x y}}=2$, find the maximum value of $x y z$. (2008 China Western Mathematical Olympiad)
38. Since $\sqrt{\frac{1-x}{y z}}+\sqrt{\frac{1-y}{z x}}+\sqrt{\frac{1-z}{x y}}=2$, by the AM-GM inequality we have
$$\begin{aligned}
2 \sqrt{x y z}= & \sqrt{x(1-x)}+\sqrt{y(1-y)}+\sqrt{z(1-z)}= \\
& \frac{1}{\sqrt{3}}[\sqrt{x(3-3 x)}+\sqrt{y(3-3 y)}+\sqrt{z(3-3 z)}] \leqslant \\
& \frac{1}{\sqrt{3}}\left[\frac{x+(3-3 x)}{2}+\frac{y+(3-3 y)}{2}+\frac{z+(3-3 z)}{2}\right]= \\
& \frac{3 \sqrt{3}}{2}-\frac{1}{\sqrt{3}}(x+y+z) \leqslant \frac{3 \sqrt{3}}{2}-\sqrt{3} \cdot \sqrt[3]{x y z}
\end{aligned}$$

Let $\sqrt[6]{x y z}=p$, then $2 p^{3} \leqslant \frac{3 \sqrt{3}}{2}-\sqrt{3} p^{2} \Leftrightarrow 4 p^{3}+2 \sqrt{3} p^{2}-3 \sqrt{3} \leqslant 0 \Leftrightarrow(2 p-\sqrt{3})$ $\left(2 p^{2}+2 \sqrt{3} p+3\right) \leqslant 0 \Leftrightarrow 2 p-\sqrt{3} \leqslant 0$, so $p \leqslant \frac{\sqrt{3}}{2}$, i.e., $x y z \leqslant \frac{2 T}{64}$, with equality holding if and only if $x=$ $y=z=\frac{3}{4}$.