Problem 993

Let $x_{1}=a=\max \left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$, $x_{1}=b: x \min \left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$.

If $x_{2} \leqslant \frac{1+b}{2}$, then $x_{1}\left(1-x_{2}\right) \geqslant x_{1}$
- $\left(1-\frac{1+b}{2}\right)=\frac{1}{2} x_{1}(1-b)$, obviously when $i=1$

H(1) holds. If $x_{2}>\frac{1+b}{2}$, since $\frac{1+b}{2} \geqslant \frac{a}{2}$, and $x_{1}=b \leqslant-1+b$, there are the following two cases:
(1) $x_{1}=b, x_{2}>\frac{1+b}{2}, x_{8}>\frac{1+b}{2}$,
$$\begin{aligned} \cdots, x_{1} & >\frac{1+b}{2} \\ x_{m} & =\min \left\{x_{2}, \cdots, x_{.}\right\}, \end{aligned}$$

where $2 \leqslant m \leqslant n$. It is easy to see that
$$x_{m-1}\left(1-x_{m}\right)>x_{1}\left(1-x_{n}\right) \text {. }$$
(2) There exists $3 \leqslant t \leqslant n$, such that $x_{t}=b \leqslant \frac{1+b}{2}$, then there exists $2 \leqslant i \leqslant n-1$ such that $x_{1}>\frac{1+v}{2}$, $x_{1+1} \leqslant \frac{1+b}{2}$, so
$$\begin{array}{l} x_{1}\left(1-x_{1+1}\right) \geqslant \frac{1+b}{2}\left(1-\frac{1+b}{2}\right) \\ \geqslant \frac{a}{4}(1-b) \geqslant \frac{1}{4} x_{1}\left(1-x_{1}\right) . \end{array}$$

In any case, (1) holds.