16. Let $a>0, x_{1}, x_{2}, \cdots, x_{n} \in[0, a](n \geqslant 2)$ and satisfy
$$x_{1} x_{2} \cdots x_{n}=\left(a-x_{1}\right)^{2}\left(a-x_{2}\right)^{2} \cdots\left(a-x_{n}\right)^{2}$$

Find the maximum value of $x_{1} x_{2} \cdots x_{n}$.
16. By the inequality of arithmetic and geometric means, we have $\left(x_{1} x_{2} \cdots x_{n}\right)^{\frac{1}{2 n}}=\left[\left(a-x_{1}\right)\left(a-x_{2}\right) \cdots(a-\right.$ $\left.\left.x_{n}\right)\right]^{\frac{1}{n}} \leqslant a-\frac{x_{1}+\cdots+x_{n}}{n} \leqslant a-\left(x_{1} \cdots x_{n}\right)^{\frac{1}{n}}$. Let $y=\left(x_{1} x_{2} \cdots x_{n}\right)^{\frac{1}{2 n}} \geqslant 0$, then we have $y \leqslant a-y^{2}$, i.e., $y^{2}+y-a \leqslant 0$. Solving the inequality gives $0 \leqslant y \leqslant \frac{-1+\sqrt{4 a+1}}{2}$, hence the maximum value of $x_{1} x_{2} \cdots x_{n}$ is $\left(\frac{-1+\sqrt{4 a+1}}{2}\right)^{2 n}$.