844

设数列 ${\displaystyle a_1,a_2,\cdots ,a_{2023}}$ 中的每项 ${\displaystyle a_i}$ 均满足 ${\displaystyle a_i\in \{-2,-1,0,1,2\}}$, 求这 ${\displaystyle 2023}$ 个数两两乘积之和的最小值.

用 ${\displaystyle b_i}$ 表示 ${\displaystyle \{a_n\}}$ 中取值为 ${\displaystyle i}$ 的值的个数，则 ${\displaystyle b_{-2}+b_{-1}+b_0+b_1+b_2=2023}$.

• ${\displaystyle b_0\ge 1}$，则$$\begin{array}{rl}\sum _{1\le i<j\le 2023}{a}_i{a}_j=& \frac{1}{2}[{\left(\sum _{i=1}^{2023}{a}_i\right)}^2-\sum _{i=1}^{2023}{a}_i^2]\\ =& \frac{1}{2}[(-2b_{-2}-b_{-1}+b_1+2b_2{)}^2-(4b_{-2}+b_{-1}+b_1+4b_2)]\\ \ge & -\frac{1}{2}(4b_{-2}+b_{-1}+b_1+4b_2)\\ \ge & -\frac{1}{2}\cdot 4(b_{-2}+b_{-1}+b_1+b_2)\ge -2\cdot 2022=-4044.\end{array}$$取等时${\displaystyle b_1=b_{-1}=0,b_0=1,b_{-2}=b_2=1011}$.
• ${\displaystyle b_0=0}$, 依照上面的放缩，$$\begin{array}{rl}\sum _{1\le i<j\le 2023}{a}_i{a}_j\ge & \frac{1}{2}(-2b_{-2}-b_{-1}+b_1+2b_2{)}^2-2(b_{-2}+b_{-1}+b_1+b_2)\\ \ge & \frac{1}{2}(-2b_{-2}-b_{-1}+b_1+2b_2{)}^2-4046.\end{array}$$
  当取值为 ${\displaystyle -4046}$, 取等条件为 $$\begin{array}{r}\{\begin{array}{rl}& -2b_{-2}-b_{-1}+b_1+2b_2=0,\\ & b_{-1}=b_1=0,\\ & b_{-2}+b_{-1}+b_1+b_2=2023,\end{array}\Rightarrow 2b_2=2023,\end{array}$$
  无解; 而 ${\displaystyle {\displaystyle \frac{1}{2}(-2b_{-2}-b_{-1}+b_1+2b_2{)}^2\ne 1}}$, 故最小值只能为 ${\displaystyle -4044}$. (此时可取 ${\displaystyle b_1=b_{-1}=b_0=0,\{b_2,b_{-2}\}=\{1011,1012\}}$).

综上，${\displaystyle {\displaystyle \sum _{1\le i<j\le 2023}{a}_i{a}_j\ge }}$ 的最小值为 ${\displaystyle -4044}$.