实数 $a_{1}, a_{2}, \cdots, a_{n}$ 满足 $a_{1}+a_{2}+\cdots+a_{n}=0$，求证：$\displaystyle \max\limits _{1 \leqslant k \leqslant n}\left(a_{k}^{2}\right) \leqslant \dfrac{n}{3} \sum\limits_{i = 1}^{n-1}\left(a_{i}-a_{i+1}\right)^{2}$
【难度】
【出处】
2006第21届CMO试题
【标注】
【答案】
略
【解析】
只须对任意 $1\leqslant k\leqslant n$，证明不等式成立即可．
记 $d_{k}=a_{k}-a_{k+1}, k=1,2, \cdots, n-1$ 则
$a_{k}=a_{k}$
$a_{k+1}=a_{k}-d_{k}, a_{k+2}=a_{k}-d_{k}-d_{k+1}, \cdots$
$a_{n}=a_{k}-d_{k}-d_{k+1}-\cdots-d_{n-1}$
$a_{k-1}=a_{k}+d_{k-1}, a_{k-2}=a_{k}+d_{k-1}+d_{k-2}, \cdots$
$a_{1}=a_{k}+d_{k-1}+d_{k-2}+\cdots+d_{1}$
把上面这 $n$ 个等式相加，并利用 $a_{1}+a_{2}+\cdots+a_{n}=0$ 可得 $n a_{k}-(n-k) d_{k}-(n-k-1) d_{k+1}-\cdots d_{n-1}+(k-1) d_{k-1}+(k-2) d_{k-2}+\cdots+d_{1}=0$
由柯西不等式可得
$\left(n a_{k}\right)^{2}=\left((n-k) d_{k}+(n-k-1) d_{k+1}+\cdots+d_{n-1}-\right.(k-1) d_{k-1}-(k-2) d_{k-2}-\cdots-d_{1} )^{2} \\\leqslant \left(\sum_{i=1}^{k=1} i^{2}+\sum_{i=1}^{n-k} i^{2}\right)\left(\sum_{i=1}^{n-1} d_{i}^{2}\right) \leqslant\left(\sum_{i=1}^{n-1} i^{2}\right)\left(\sum_{i=1}^{n-1} d_{i}^{2}\right)=\frac{n(n-1)(2 n-1)}{6}\left(\sum_{i=1}^{n-1} d_{i}^{2}\right) \leqslant \frac{n^{3}}{3}\left(\sum_{i=1}^{n-1} d_{i}^{2}\right) $
所以 $\displaystyle a_{k}^{2} \leqslant \frac{n}{3} \sum\limits_{i=1}^{n-1}\left(a_{i}-a_{i+1}\right)^{2}$
记 $d_{k}=a_{k}-a_{k+1}, k=1,2, \cdots, n-1$ 则
$a_{k}=a_{k}$
$a_{k+1}=a_{k}-d_{k}, a_{k+2}=a_{k}-d_{k}-d_{k+1}, \cdots$
$a_{n}=a_{k}-d_{k}-d_{k+1}-\cdots-d_{n-1}$
$a_{k-1}=a_{k}+d_{k-1}, a_{k-2}=a_{k}+d_{k-1}+d_{k-2}, \cdots$
$a_{1}=a_{k}+d_{k-1}+d_{k-2}+\cdots+d_{1}$
把上面这 $n$ 个等式相加，并利用 $a_{1}+a_{2}+\cdots+a_{n}=0$ 可得 $n a_{k}-(n-k) d_{k}-(n-k-1) d_{k+1}-\cdots d_{n-1}+(k-1) d_{k-1}+(k-2) d_{k-2}+\cdots+d_{1}=0$
由柯西不等式可得
$\left(n a_{k}\right)^{2}=\left((n-k) d_{k}+(n-k-1) d_{k+1}+\cdots+d_{n-1}-\right.(k-1) d_{k-1}-(k-2) d_{k-2}-\cdots-d_{1} )^{2} \\\leqslant \left(\sum_{i=1}^{k=1} i^{2}+\sum_{i=1}^{n-k} i^{2}\right)\left(\sum_{i=1}^{n-1} d_{i}^{2}\right) \leqslant\left(\sum_{i=1}^{n-1} i^{2}\right)\left(\sum_{i=1}^{n-1} d_{i}^{2}\right)=\frac{n(n-1)(2 n-1)}{6}\left(\sum_{i=1}^{n-1} d_{i}^{2}\right) \leqslant \frac{n^{3}}{3}\left(\sum_{i=1}^{n-1} d_{i}^{2}\right) $
所以 $\displaystyle a_{k}^{2} \leqslant \frac{n}{3} \sum\limits_{i=1}^{n-1}\left(a_{i}-a_{i+1}\right)^{2}$
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