844

844

设数列 $a_{1}, a_{2}, \dots, a_{2023}$ 中的每项 $a_{i}$ 均满足 $a_{i} \in {- 2, - 1, 0, 1, 2}$ , 求这 $2023$ 个数两两乘积之和的最小值.

解钱莘芪

用 $b_{i}$ 表示 ${a_{n}}$ 中取值为 $i$ 的值的个数，则 $b_{- 2} + b_{- 1} + b_{0} + b_{1} + b_{2} = 2023$ .

$b_{0} \geq 1$ ，则 $\begin{aligned} \sum_{1 \leq i < j \leq 2023} a_{i} a_{j} = & \frac{1}{2} [{(\sum_{i = 1}^{2023} a_{i})}^{2} - \sum_{i = 1}^{2023} a_{i}^{2}] = \frac{1}{2} [(- 2 b_{- 2} - b_{- 1} + b_{1} + 2 b_{2})^{2} - (4 b_{- 2} + b_{- 1} + b_{1} + 4 b_{2})] \\ \geq & - \frac{1}{2} (4 b_{- 2} + b_{- 1} + b_{1} + 4 b_{2}) \geq - \frac{1}{2} \cdot 4 (b_{- 2} + b_{- 1} + b_{1} + b_{2}) \geq - 2 \cdot 2022 = - 4044. \end{aligned}$ 取等时 $b_{1} = b_{- 1} = 0, b_{0} = 1, b_{- 2} = b_{2} = 1011$ .
$b_{0} = 0$ , 依照上面的放缩， $\begin{aligned} \sum_{1 \leq i < j \leq 2023} a_{i} a_{j} \geq & \frac{1}{2} (- 2 b_{- 2} - b_{- 1} + b_{1} + 2 b_{2})^{2} - 2 (b_{- 2} + b_{- 1} + b_{1} + b_{2}) \\ \geq & \frac{1}{2} (- 2 b_{- 2} - b_{- 1} + b_{1} + 2 b_{2})^{2} - 4046. \end{aligned}$
当取值为 $- 4046$ , 取等条件为 $\begin{array}{r} {\begin{aligned} - 2 b_{- 2} - b_{- 1} + b_{1} + 2 b_{2} = 0, \\ b_{- 1} = b_{1} = 0, \\ b_{- 2} + b_{- 1} + b_{1} + b_{2} = 2023, \end{aligned} \Rightarrow 2 b_{2} = 2023, \end{array}$
无解; 而 $\frac{1}{2} (- 2 b_{- 2} - b_{- 1} + b_{1} + 2 b_{2})^{2} \neq 1$ , 故最小值只能为 $- 4044$ . (此时可取 $b_{1} = b_{- 1} = b_{0} = 0, {b_{2}, b_{- 2}} = {1011, 1012}$ ). 综上， $\sum_{1 \leq i < j \leq 2023} a_{i} a_{j} \geq$ 的最小值为 $- 4044$ .