直面心魔——广东预赛二试回顾(其一)

\(2025\)广东预赛已经告一段落,但是糖丸的二试还是令我念念不忘,故在此直面心魔.

先看代数题. ## 题目 已知非负实数\(x_1,x_2,\cdots,x_n\)满足:\(\displaystyle \sum_i^5x_i^2=4\).设$$S=\sum_i^5\frac1{1+x_i}\sum_i^5x_i.$$求\(S\)的最大值.

分析

猜取等

\(x_i\)	\(S\)
全取\(\displaystyle \frac{2\sqrt5}{5}\)	\(\displaystyle 50(\sqrt5-2)\)
\(1,1,1,1,0\)	\(12\)
\(2,0,0,0,0\)	\(\displaystyle \frac{26}3\)

猜测是\(1,1,1,1,0\).

凑取等

因为想要在\(0,1\)处都取等,我们选用\(\begin{equation}x\left(x-1\right)^2\geqslant0\end{equation}\)进行放缩.对\((1)\)进行恒等变形得到\(\begin{equation}\frac14x^2-\frac34x+1\geqslant\frac1{1+x}.\end{equation}\)这正是我们想要的.

解答

由\(x\geqslant0\),得$$x\left(x-1\right)^2\geqslant0 \Leftrightarrow \frac14x^2-\frac34x+1\geqslant\frac1{1+x}.$$求和得到$$\begin{align*}S&=\sum_i^5\frac1{1+x_i}\sum_i^5x_i\\\\&\leqslant\sum_i^5\left(\frac14x^2-\frac34x+1\right)\sum_i^5x_i\\\\&=\frac34\left(8-\sum_i^5x_i\right)\sum_i^5x_i\\\\&\leqslant\frac34\left(\frac82\right)^2=12.\end{align*}$$等号成立当且仅当\(x_i\)取\(4\)个\(1\),\(1\)个\(0\).\(\square\)