设 $a_{1}, a_{2}, \cdots, a_{n}$ 是 $n$ 个非负实数，记 $\displaystyle S_{k}=\sum\limits_{i=1}^{k} a_{i}, 1 \leqslant k \leqslant n$．证明：$\displaystyle \sum\limits_{i=1}^{n}\left(a_{i} S_{i} \sum_{j=i}^{n} a_{j}^{2}\right) \leqslant \sum_{i=1}^{n}\left(a_{i} S_{i}\right)^{2}$．
【难度】
【出处】
2016年中国西部数学邀请赛试题
【标注】
【答案】
略
【解析】
证法一
令 $\displaystyle b_{i}=a_{i} S_{i}, c_{i}=\sum\limits_{j=i}^{n} a_{j}^{2}, i=1.2, \cdots, n$．
则原不等式等价于 $\displaystyle \sum\limits_{i=1}^{n} b_{i} c_{i} \leqslant \sum_{i=1}^{n} b_{i}^{2}$ ①
注意到对 $1 \leqslant i \leqslant n$，有
$\begin{aligned} B_{i} &=b_{1}+b_{2}+\cdots+b_{i} \\ &=a_{1} S_{1}+a_{2} S_{2}+\cdots+a_{i} S_{i} \\ & \leqslant\left(a_{1}+a_{2}+\cdots+a_{i}\right) S_{i} \\ &=S_{i}^{2} \end{aligned}$
故由阿贝尔恒等式有
$\begin{aligned} \sum_{i=1}^{n} b_{i} c_{i} &=\sum_{i=1}^{n-1} B_{i}\left(c_{i}-c_{i+1}\right)+B_{n} c_{n} \\ & \leqslant \sum_{i=1}^{n-1} a_{i} S_{i}^{2}+B_{n} c_{n} \\ & \leqslant \sum_{i=1}^{n} a_{i} S_{i}^{2}=\sum_{i=1}^{n} b_{i}^{2} \end{aligned}$
故 ① 式成立，所以原不等式成立．
证法二
注意到下面的恒等式
$\displaystyle \sum\limits_{i=1}^{n}\left(a_{i} S_{i} \sum_{j=i}^{n} a_{j}^{2}\right)=\sum_{j=1}^{n}\left(a_{j}^{2} \sum_{i=1}^{j} a_{i} S_{i}\right)$
故要证明原不等式，只需证明 $\displaystyle \sum\limits_{j=1}^{n}\left(a_{j}^{2} \sum_{i=1}^{j} a_i S_{i}\right) \leqslant \sum_{j=1}^{n} a_{j}^{2} S_{i}^{2}$
对 $1\leqslant j\leqslant n$，比较上式两端 $a_j^2$ 的系数，要使得上式成立，只需 $\displaystyle \sum\limits_{i=1}^{j} a_iS_{i} \leqslant S_{j}^{2}$
事实上，$\displaystyle \sum\limits_{i=1}^{j} a_i S_{i} \leqslant\left(\sum_{i=1}^{j} a_{i}\right) S_{j}=S_{j}^{2}$．故上式成立．
所以，原不等式成立．
令 $\displaystyle b_{i}=a_{i} S_{i}, c_{i}=\sum\limits_{j=i}^{n} a_{j}^{2}, i=1.2, \cdots, n$．
则原不等式等价于 $\displaystyle \sum\limits_{i=1}^{n} b_{i} c_{i} \leqslant \sum_{i=1}^{n} b_{i}^{2}$ ①
注意到对 $1 \leqslant i \leqslant n$，有
$\begin{aligned} B_{i} &=b_{1}+b_{2}+\cdots+b_{i} \\ &=a_{1} S_{1}+a_{2} S_{2}+\cdots+a_{i} S_{i} \\ & \leqslant\left(a_{1}+a_{2}+\cdots+a_{i}\right) S_{i} \\ &=S_{i}^{2} \end{aligned}$
故由阿贝尔恒等式有
$\begin{aligned} \sum_{i=1}^{n} b_{i} c_{i} &=\sum_{i=1}^{n-1} B_{i}\left(c_{i}-c_{i+1}\right)+B_{n} c_{n} \\ & \leqslant \sum_{i=1}^{n-1} a_{i} S_{i}^{2}+B_{n} c_{n} \\ & \leqslant \sum_{i=1}^{n} a_{i} S_{i}^{2}=\sum_{i=1}^{n} b_{i}^{2} \end{aligned}$
故 ① 式成立，所以原不等式成立．
证法二
注意到下面的恒等式
$\displaystyle \sum\limits_{i=1}^{n}\left(a_{i} S_{i} \sum_{j=i}^{n} a_{j}^{2}\right)=\sum_{j=1}^{n}\left(a_{j}^{2} \sum_{i=1}^{j} a_{i} S_{i}\right)$
故要证明原不等式，只需证明 $\displaystyle \sum\limits_{j=1}^{n}\left(a_{j}^{2} \sum_{i=1}^{j} a_i S_{i}\right) \leqslant \sum_{j=1}^{n} a_{j}^{2} S_{i}^{2}$
对 $1\leqslant j\leqslant n$，比较上式两端 $a_j^2$ 的系数，要使得上式成立，只需 $\displaystyle \sum\limits_{i=1}^{j} a_iS_{i} \leqslant S_{j}^{2}$
事实上，$\displaystyle \sum\limits_{i=1}^{j} a_i S_{i} \leqslant\left(\sum_{i=1}^{j} a_{i}\right) S_{j}=S_{j}^{2}$．故上式成立．
所以，原不等式成立．
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