$$\begin{aligned} & \sqrt[k]{x_1^k + x_n^k - \frac{x_1^k + \dots + x_n^k}{n}} = \sqrt[k]{\frac{(x_1^k + x_n^k - x_1^k) + \dots + (x_1^k + x_n^k - x_n^k)}{n}} \\ & k \text{ th mean} \ge \frac{\sqrt[k]{x_1^k + x_n^k - x_1^k} + \sqrt[k]{x_1^k + x_n^k - x_2^k} + \dots + \sqrt[k]{x_1^k + x_n^k - x_n^k}}{n} \\ & \text{Since } x_1+x_n - \frac{x_1 + \dots + x_n}{n} = \frac{(x_1+x_n-x_1) + (x_1+x_n-x_2) + \dots + (x_1+x_n-x_n)}{n} \\ & \text{it is sufficient to prove that } \sqrt[k]{x_1^k + x_n^k - x_i^k} \ge x_1 + x_n - x_i, \forall i = 1, n. \end{aligned}$$
Note that $\sqrt[k]{x_1^k + x_n^k - x_i^k} \ge x_1 + x_n - x_i \Leftrightarrow x_1^k + x_n^k - x_i^k \ge (x_1 + x_n - x_i)^k \Leftrightarrow x_1^k + x_n^k \ge x_i^k + (x_1 + x_n - x_i)^k$. Substituting $x_1 = a, x_i = a + x, x_n = a + x + y, x, y \ge 0$ we get $a^k + (a+x+y)^k \ge (a+x)^k + (a+y)^k \Leftrightarrow 2a^k + \sum_{i=1}^k (x+y)^i a^{k-i} C_k^i \ge 2a^k + \sum_{i=1}^k (x^i + y^i)a^{k-i} C_k^i \Leftrightarrow \sum_{i=1}^k ((x+y)^i - x^i - y^i)a^{k-i} C_k^i \ge 0$ and we have proved. Equality holds when $x = y = 0 \Rightarrow x_1 = x_2 = \dots = x_n$.