(2007浙江省赛B卷最后一题)设$sumlimits_{i=1}^{n}{x_i}=1,x_i>0,$求证:$nsumlimits_{i=1}^n{x_i^2}-sumlimits_{i<j}{dfrac{(x_i-x_j)^2}{x_i+x_j}}le1$
证明:
egin{align*}
& nsumlimits_{i=1}^n{x_i^2}-sumlimits_{i<j}{dfrac{(x_i-x_j)^2}{x_i+x_j}} \
&=(sumlimits_{i=1}^n{x_i})^2+sumlimits_{i<j}{(x_i-x_j)^2}-sumlimits_{i<j}{dfrac{(x_i-x_j)^2}{x_i+x_j}}\
&=1+sumlimits_{i<j}{dfrac{(x_i-x_j)^2(x_i+x_j-1)}{x_i+x_j}}le1
end{align*}