给定整数 $n\geqslant 3$，实数 $a_{1}, a_{2}, \cdots, a_{n}$ 满足 $\min\limits _{1 \leqslant i<j \leqslant n} | a_{i}-a_{j} |=1$，求 $\displaystyle \sum\limits_{k=1}^{n}\left|a_{k}\right|^{3}$ 的最小值．
【难度】
【出处】
2009第24届CMO试题
【标注】
【答案】
略
【解析】
不妨设 $a_{1}<a_{2}<\cdots<a_{n}$，则对 $1 \leqslant k \leqslant n$，有 $\left|a_{k}\right|+\left|a_{n-k+1}\right| \geqslant\left|a_{n-k+1}-a_{k}\right| \geqslant|n+1-2 k|$ 所以
$\displaystyle \sum\limits_{k=1}^{n}\left|a_{k}\right|^{3}=\frac{1}{2} \sum_{k=1}^{n}\left(\left|a_{k}\right|^{3}+\left|a_{n+1-k}\right|^{3}\right)=\\ \frac{1}{2} \sum\limits_{k=1}^{n}\left(\left|a_{k}\right|+\left|a_{n+1-k}\right|\right)\left(\frac{3}{4}\left(\left|a_{k}\right|- \left|a_{n+1-k}\right| \right)^{2}+\frac{1}{4}\left(\left|a_{k}\right|+\left|a_{n+1-k}\right|\right)^{2} \right)\\ \geqslant \frac{1}{8} \sum\limits_{k=1}^{n}\left(\left|a_{k}\right|+\left|a_{n+1-k}\right|\right)^{3} \geqslant \frac{1}{8} \sum\limits_{k=1}^{n}|n+1-2 k|^{3} $
当 $n$ 为奇数时 $\displaystyle \sum\limits_{k=1}^{n} | n+1-2\left.k\right|^{3}=2 \cdot 2^{3} \cdot \sum_{i=1}^{\frac{n-1}{2}} i^{3}=\frac{1}{4}\left(n^{2}-1\right)^{2}$
当 $n$ 为偶数时 $\displaystyle \begin{aligned} \sum\limits_{k=1}^{n}|n+1-2 k|^{3}=& 2 \sum_{i=1}^{\frac{n}{2}}(2 i-1)^{3}=2\left(\sum_{j=1}^{n} j^{3}-\sum_{i=1}^{\frac{n}{2}}(2 i)^{3}\right)= \frac{1}{4} n^{2}\left(n^{2}-2\right) \end{aligned}$
所以，当 $n$ 为奇数时，$\displaystyle \sum\limits_{k=1}^{n}\left|a_{k}\right|^{3} \geqslant \frac{1}{32}\left(n^{2}-1\right)^{2}$，当 $n$ 为偶数时，$\displaystyle \sum\limits_{k=1}^{n}\left|a_{k}\right|^{3} \geqslant \dfrac{1}{32} n^{2}\left(n^{2}-2\right)$，等号均在 $a_{i}=i-\dfrac{n+1}{2}, i=1,2, \cdots, n$ 时成立．
因此，$\displaystyle \sum\limits_{k=1}^{n}\left|a_{k}\right|^{3}$ 的最小值为 $\dfrac{1}{32}\left(n^{2}-1\right)^{2}$（$n$ 为奇数），或者 $\dfrac{1}{32} n^{2}\left(n^{2}-2\right)$（$n$ 为偶数）．
$\displaystyle \sum\limits_{k=1}^{n}\left|a_{k}\right|^{3}=\frac{1}{2} \sum_{k=1}^{n}\left(\left|a_{k}\right|^{3}+\left|a_{n+1-k}\right|^{3}\right)=\\ \frac{1}{2} \sum\limits_{k=1}^{n}\left(\left|a_{k}\right|+\left|a_{n+1-k}\right|\right)\left(\frac{3}{4}\left(\left|a_{k}\right|- \left|a_{n+1-k}\right| \right)^{2}+\frac{1}{4}\left(\left|a_{k}\right|+\left|a_{n+1-k}\right|\right)^{2} \right)\\ \geqslant \frac{1}{8} \sum\limits_{k=1}^{n}\left(\left|a_{k}\right|+\left|a_{n+1-k}\right|\right)^{3} \geqslant \frac{1}{8} \sum\limits_{k=1}^{n}|n+1-2 k|^{3} $
当 $n$ 为奇数时 $\displaystyle \sum\limits_{k=1}^{n} | n+1-2\left.k\right|^{3}=2 \cdot 2^{3} \cdot \sum_{i=1}^{\frac{n-1}{2}} i^{3}=\frac{1}{4}\left(n^{2}-1\right)^{2}$
当 $n$ 为偶数时 $\displaystyle \begin{aligned} \sum\limits_{k=1}^{n}|n+1-2 k|^{3}=& 2 \sum_{i=1}^{\frac{n}{2}}(2 i-1)^{3}=2\left(\sum_{j=1}^{n} j^{3}-\sum_{i=1}^{\frac{n}{2}}(2 i)^{3}\right)= \frac{1}{4} n^{2}\left(n^{2}-2\right) \end{aligned}$
所以，当 $n$ 为奇数时，$\displaystyle \sum\limits_{k=1}^{n}\left|a_{k}\right|^{3} \geqslant \frac{1}{32}\left(n^{2}-1\right)^{2}$，当 $n$ 为偶数时，$\displaystyle \sum\limits_{k=1}^{n}\left|a_{k}\right|^{3} \geqslant \dfrac{1}{32} n^{2}\left(n^{2}-2\right)$，等号均在 $a_{i}=i-\dfrac{n+1}{2}, i=1,2, \cdots, n$ 时成立．
因此，$\displaystyle \sum\limits_{k=1}^{n}\left|a_{k}\right|^{3}$ 的最小值为 $\dfrac{1}{32}\left(n^{2}-1\right)^{2}$（$n$ 为奇数），或者 $\dfrac{1}{32} n^{2}\left(n^{2}-2\right)$（$n$ 为偶数）．
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