Jensen's inequality for the square function

Expansion. Compute
$$\sum_{i<j}(x_i-x_j)^2=\sum_{i<j}(x_i^2+x_j^2-2x_ix_j).$$
Each $x_i^2$ appears $n-1$ times, so
$$\sum_{i<j}(x_i-x_j)^2=(n-1)\sum_i x_i^2-2\sum_{i<j}x_ix_j.$$
Now $(n-1)\sum x_i^2= n\sum x_i^2-\sum x_i^2$ and $2\sum_{i<j}x_ix_j=(\sum x_i)^2-\sum x_i^2$, so
$$\sum_{i<j}(x_i-x_j)^2= n\sum x_i^2-(\sum x_i)^2.$$

Inequality. Since the LHS is $\ge0$, we get $n\sum x_i^2\ge(\sum x_i)^2$, i.e.
$$\frac{x_1^2+\cdots+x_n^2}{n}\ge\left(\frac{x_1+\cdots+x_n}{n}\right)^2.$$

Conclusion. The square function is convex, so Jensen's inequality applies.