MT【194】又见和式变换

(2007浙江省赛B卷最后一题)设$\sum\limits_{i=1}^{n}{x_i}=1,x_i>0,$求证:$n\sum\limits_{i=1}^n{x_i^2}-\sum\limits_{i<j}{\dfrac{(x_i-x_j)^2}{x_i+x_j}}\le1$

证明:
$$\begin{align*} & n\sum\limits_{i=1}^n{x_i^2}-\sum\limits_{i<j}{\dfrac{(x_i-x_j)^2}{x_i+x_j}} \\ &=(\sum\limits_{i=1}^n{x_i})^2+\sum\limits_{i<j}{(x_i-x_j)^2}-\sum\limits_{i<j}{\dfrac{(x_i-x_j)^2}{x_i+x_j}}\\ &=1+\sum\limits_{i<j}{\dfrac{(x_i-x_j)^2(x_i+x_j-1)}{x_i+x_j}}\le1 \end{align*}$$