Mongolian National Mathematical Olympiad

Replacing $x_i$ by $x_i-x_n$ we can assume that $x_n=0$. Since the left hand side of the inequality is $2(\sum x_i^2)-2(2n-1)A^2$, it is enough to show that $\sum x_i^2\ge 2(2n-1)A^2$. By the Cauchy-Schwarz inequality, we have $$\frac{2n-1}{n-1}\left(\frac{{\sum }_{i=1}^{n-1}{x}_i^2}{n-1}\right)\ge 2{\left(\frac{{\sum }_{i=1}^{n-1}{x}_i}{n-1}\right)}^2.$$ The expression $(2n-1)(\sum x_i^2)/(n-1{)}^2$ is greater than the left side while the right side is greater than $2((2n-1)A/(n-1){)}^2$ since $x_1\le x_2\le \cdots \le x_n=0$. Equality holds if and only if $x_1=\cdots =x_{2n-1}$.