Mongolian Mathematical Olympiad

$$\begin{array}{rl}& \sqrt[k]{x_1^k+x_n^k-\frac{x_1^k+\cdots +x_n^k}{n}}=\sqrt[k]{\frac{(x_1^k+x_n^k-x_1^k)+\cdots +(x_1^k+x_n^k-x_n^k)}{n}}\\ & k\text{ th mean}\ge \frac{\sqrt[k]{x_1^k+x_n^k-x_1^k}+\sqrt[k]{x_1^k+x_n^k-x_2^k}+\cdots +\sqrt[k]{x_1^k+x_n^k-x_n^k}}{n}\\ & \text{Since }{x}_1+x_n-\frac{x_1+\cdots +x_n}{n}=\frac{(x_1+x_n-x_1)+(x_1+x_n-x_2)+\cdots +(x_1+x_n-x_n)}{n}\\ & \text{it is sufficient to prove that }\sqrt[k]{x_1^k+x_n^k-x_i^k}\ge x_1+x_n-x_i,\forall i=1,n.\end{array}$$ Note that $\sqrt[k]{x_1^k+x_n^k-x_i^k}\ge x_1+x_n-x_i\iff x_1^k+x_n^k-x_i^k\ge (x_1+x_n-x_i{)}^k\iff x_1^k+x_n^k\ge x_i^k+(x_1+x_n-x_i{)}^k$. Substituting $x_1=a,x_i=a+x,x_n=a+x+y,x,y\ge 0$ we get $a^k+(a+x+y{)}^k\ge (a+x{)}^k+(a+y{)}^k\iff 2a^k+{\sum }_{i=1}^k(x+y{)}^i{a}^{k-i}{C}_k^i\ge 2a^k+{\sum }_{i=1}^k(x^i+y^i)a^{k-i}{C}_k^i\iff {\sum }_{i=1}^k((x+y{)}^i-x^i-y^i)a^{k-i}{C}_k^i\ge 0$ and we have proved. Equality holds when $x=y=0\Rightarrow x_1=x_2=\cdots =x_n$.