设整数 $n\geqslant 2$，证明：对任意正实数 $a_{1}, a_{2}, \cdots, a_{n}$，都有 $\displaystyle \sum\limits_{i=1}^{n} \max \left\{a_{1}, a_{2}, \cdots, a_{i}\right\} \cdot \min \left\{a_{i}, a_{i+1}, \cdots, a_{n}\right\} \leqslant \dfrac{n}{2 \sqrt{n-1}} \cdot \sum_{i=1}^{n} a_{i}^{2}$．
【难度】
【出处】
2017年中国西部数学邀请赛试题
【标注】
【答案】
略
【解析】
对 $n$ 用第二数学归纳法．
当 $n=2$ 时，左式 $=a_{1} \cdot \min \left\{a_{1}, a_{2}\right\}+\max \left\{a_{1}, a_{2}\right\} \cdot a_{2}$．
若 $a_1\geqslant a_2$，则原式等价于 $2 a_{1} a_{2} \leqslant a_{1}^{2}+a_{2}^{2}$，命题成立；
若 $a_1\leqslant a_2$，则原式等价于 $a_{1}^{2}+a_{2}^{2} \leqslant a_{1}^{2}+a_{2}^{2}$，命题成立；
假设命题对所有大于等于 $2$ 且小于 $n$ 的正整数成立，来看 $n$ 时的情形．
对 $2\leqslant i\leqslant n$，记 $c_{i}=\dfrac{i}{2 \sqrt{i-1}}$，再定义 $c_1= 1$，容易验证 $c_{1}=c_{2}<c_{3}<\cdots<c_{n}$．
记 $M=\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$，并设 $a_{k}=M$．
当 $k=1$ 时，原式 $\displaystyle =M \sum\limits_{i=1}^{n} \min \left\{a_{i}, a_{i+1}, \cdots, a_{n}\right\}$，因为 $\displaystyle \min \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}=\min \left\{a_{2}, \cdots, a_{n}\right\} \leqslant \dfrac{1}{n-1} \sum\limits_{i=2}^{n} a_{i}$
且当 $2 \leqslant i \leqslant n$ 时，$\min \left\{a_{i}, a_{i+1}, \cdots, a_{n}\right\} \leqslant a_{i}$，所以 $\displaystyle \sum\limits_{i=1}^{n} \min \left\{a_{i}, a_{i+1}, \cdots, a_{n}\right\} \leqslant \frac{1}{n-1} \sum_{i=2}^{n} a_{i}+\sum_{i=2}^{n} a_{i}=\frac{n}{n-1} \sum_{i=2}^{n} a_{i}$．
由均值不等式，
原式 $\displaystyle \leqslant \dfrac{n}{n-1} M \sum\limits_{i=2}^{n} a_{i} \leqslant \dfrac{n}{2 \sqrt{n-1}}\left[M^{2}+\frac{1}{n-1}\left(\sum_{i=2}^{n} a_{i}\right)^{2}\right] \leqslant \dfrac{n}{2 \sqrt{n-1}}\left(M^{2}+\sum_{i=2}^{n} a_{i}^{2}\right)=\dfrac{n}{2 \sqrt{n-1}} \sum_{i=1}^{n} a_{i}^{2}$
当 $k=n$ 时，$\min \left\{a_{i}, a_{*+1}, \cdots, a_{n}\right\}=\min \left\{a_{i}, a_{i+1}, \cdots, a_{n-1}\right\}$，所以
原式 $=\sum_{i=1}^{n-1} \max \left\{a_{1}, a_{2}, \cdots, a_{i}\right\} \cdot \min \left\{a_{i}, a_{i+1}, \cdots, a_{n-1}\right\}+M^{2}$，
由归纳假设，$\displaystyle \sum\limits_{i=1}^{n-1} \max \left\{a_{1}, a_{2}, \cdots, a_{i}\right\} \cdot \min \left\{a_{i}, a_{i+1}, \cdots, a_{n-1}\right\} \leqslant c_{n-1} \sum_{i=1}^{n-1} a_{i}^{2}$．
所以，原式 $\displaystyle \leqslant c_{n-1} \sum\limits_{i=1}^{n-1} a_{i}^{2}+M^{2}<\dfrac{n}{2 \sqrt{n-1}}\left(\sum_{i=1}^{n-1} a_{i}^{2}+M^{2}\right)=\dfrac{n}{2 \sqrt{n-1}} \sum_{i=1}^{n} a_{i}^{2}$．
当 $2\leqslant k\leqslant n-1$ 时，结合 $k=1$ 和 $k=n$ 时的证明得，
原式 $\displaystyle =\sum\limits_{i=1}^{k-1} \max \left\{a_{1}, a_{2}, \cdots, a_{i}\right\} \cdot \min \left\{a_{i}, a_{i+1}, \cdots, a_{n}\right\}+M \sum_{i=k}^{n} \min \left\{a_{i}, a_{i+1}, \cdots, a_{n}\right\}$
$\begin{aligned} & \leqslant \sum_{i=1}^{k-1} \max \left\{a_{1}, a_{2}, \cdots, a_{i}\right\} \cdot \min \left\{a_{i}, a_{i+1}, \cdots, a_{k-1}\right\}+\dfrac{n-k+1}{n-k} M \sum_{i=k+1}^{n}a_i \\ & \leqslant c_{k-1} \sum_{i=1}^{k-1} a_{i}^{2}+\dfrac{n-k+1}{2 \sqrt{n-k}}\left(M^{2}+\sum_{i=k+1}^{n} a_{i}^{2}\right) =c_{k-1} \sum_{i=1}^{k-1} a_{i}^{2}+c_{n-k+1} \sum_{i=k}^{n} a_{i}^{2}<\dfrac{n}{2 \sqrt{n-1}} \sum_{i=1}^{n} a_{i}^{2} \end{aligned}$
综上，命题得证．
当 $n=2$ 时，左式 $=a_{1} \cdot \min \left\{a_{1}, a_{2}\right\}+\max \left\{a_{1}, a_{2}\right\} \cdot a_{2}$．
若 $a_1\geqslant a_2$，则原式等价于 $2 a_{1} a_{2} \leqslant a_{1}^{2}+a_{2}^{2}$，命题成立；
若 $a_1\leqslant a_2$，则原式等价于 $a_{1}^{2}+a_{2}^{2} \leqslant a_{1}^{2}+a_{2}^{2}$，命题成立；
假设命题对所有大于等于 $2$ 且小于 $n$ 的正整数成立，来看 $n$ 时的情形．
对 $2\leqslant i\leqslant n$，记 $c_{i}=\dfrac{i}{2 \sqrt{i-1}}$，再定义 $c_1= 1$，容易验证 $c_{1}=c_{2}<c_{3}<\cdots<c_{n}$．
记 $M=\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$，并设 $a_{k}=M$．
当 $k=1$ 时，原式 $\displaystyle =M \sum\limits_{i=1}^{n} \min \left\{a_{i}, a_{i+1}, \cdots, a_{n}\right\}$，因为 $\displaystyle \min \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}=\min \left\{a_{2}, \cdots, a_{n}\right\} \leqslant \dfrac{1}{n-1} \sum\limits_{i=2}^{n} a_{i}$
且当 $2 \leqslant i \leqslant n$ 时，$\min \left\{a_{i}, a_{i+1}, \cdots, a_{n}\right\} \leqslant a_{i}$，所以 $\displaystyle \sum\limits_{i=1}^{n} \min \left\{a_{i}, a_{i+1}, \cdots, a_{n}\right\} \leqslant \frac{1}{n-1} \sum_{i=2}^{n} a_{i}+\sum_{i=2}^{n} a_{i}=\frac{n}{n-1} \sum_{i=2}^{n} a_{i}$．
由均值不等式，
原式 $\displaystyle \leqslant \dfrac{n}{n-1} M \sum\limits_{i=2}^{n} a_{i} \leqslant \dfrac{n}{2 \sqrt{n-1}}\left[M^{2}+\frac{1}{n-1}\left(\sum_{i=2}^{n} a_{i}\right)^{2}\right] \leqslant \dfrac{n}{2 \sqrt{n-1}}\left(M^{2}+\sum_{i=2}^{n} a_{i}^{2}\right)=\dfrac{n}{2 \sqrt{n-1}} \sum_{i=1}^{n} a_{i}^{2}$
当 $k=n$ 时，$\min \left\{a_{i}, a_{*+1}, \cdots, a_{n}\right\}=\min \left\{a_{i}, a_{i+1}, \cdots, a_{n-1}\right\}$，所以
原式 $=\sum_{i=1}^{n-1} \max \left\{a_{1}, a_{2}, \cdots, a_{i}\right\} \cdot \min \left\{a_{i}, a_{i+1}, \cdots, a_{n-1}\right\}+M^{2}$，
由归纳假设，$\displaystyle \sum\limits_{i=1}^{n-1} \max \left\{a_{1}, a_{2}, \cdots, a_{i}\right\} \cdot \min \left\{a_{i}, a_{i+1}, \cdots, a_{n-1}\right\} \leqslant c_{n-1} \sum_{i=1}^{n-1} a_{i}^{2}$．
所以，原式 $\displaystyle \leqslant c_{n-1} \sum\limits_{i=1}^{n-1} a_{i}^{2}+M^{2}<\dfrac{n}{2 \sqrt{n-1}}\left(\sum_{i=1}^{n-1} a_{i}^{2}+M^{2}\right)=\dfrac{n}{2 \sqrt{n-1}} \sum_{i=1}^{n} a_{i}^{2}$．
当 $2\leqslant k\leqslant n-1$ 时，结合 $k=1$ 和 $k=n$ 时的证明得，
原式 $\displaystyle =\sum\limits_{i=1}^{k-1} \max \left\{a_{1}, a_{2}, \cdots, a_{i}\right\} \cdot \min \left\{a_{i}, a_{i+1}, \cdots, a_{n}\right\}+M \sum_{i=k}^{n} \min \left\{a_{i}, a_{i+1}, \cdots, a_{n}\right\}$
$\begin{aligned} & \leqslant \sum_{i=1}^{k-1} \max \left\{a_{1}, a_{2}, \cdots, a_{i}\right\} \cdot \min \left\{a_{i}, a_{i+1}, \cdots, a_{k-1}\right\}+\dfrac{n-k+1}{n-k} M \sum_{i=k+1}^{n}a_i \\ & \leqslant c_{k-1} \sum_{i=1}^{k-1} a_{i}^{2}+\dfrac{n-k+1}{2 \sqrt{n-k}}\left(M^{2}+\sum_{i=k+1}^{n} a_{i}^{2}\right) =c_{k-1} \sum_{i=1}^{k-1} a_{i}^{2}+c_{n-k+1} \sum_{i=k}^{n} a_{i}^{2}<\dfrac{n}{2 \sqrt{n-1}} \sum_{i=1}^{n} a_{i}^{2} \end{aligned}$
综上，命题得证．
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