China Western Mathematical Olympiad

Denote $u=\sqrt[6]{xyz}$. Then by the given condition and mean inequality, $$\begin{array}{rl}2u^3& =2\sqrt{xyz}=\frac{1}{\sqrt{3}}\sum \sqrt{x(3-3x)}\\ & \le \frac{1}{\sqrt{3}}\sum \frac{x+(3-3x)}{2}=\frac{3\sqrt{3}}{2}-\frac{1}{\sqrt{3}}(x+y+z)\\ & \le \frac{3\sqrt{3}}{2}-\sqrt{3}\times \sqrt[3]{xyz}=\frac{3\sqrt{3}}{2}-\sqrt{3}{u}^2.\end{array}$$ Therefore, $4u^3+2\sqrt{3}{u}^2-3\sqrt{3}\le 0$, i.e. $$(2u-\sqrt{3})(2u^2+2\sqrt{3}u+3)\le 0,$$ and thus $u\le \frac{\sqrt{3}}{2}$. Following this, we have $xyz\le \frac{27}{64}$, and equality holds when $x=y=z=\frac{3}{4}$. Hence, the maximum is $\frac{27}{64}$.