Show that the inequality

holds for all real numbers $x_1,x_2,\dots ,x_n$.

Step 1 (Reduction): WLOG, we can add both $x_i$ and $-x_i$ to the list (replacing $n$ by $2n$) and the inequality becomes equivalent. This symmetrization trick transforms the problem.

Step 2 (Key identity): For real numbers $a,b$: $\sqrt{|a+b|}+\sqrt{|a-b|}\ge 2\sqrt{|a|}$ by the concavity of $\sqrt{\cdot }$ ... actually we need the opposite direction.

Step 3 (Integral representation): Use the identity $\sqrt{|t|}=c{\int }_0^{\infty }\frac{1-\cos (\xi t)}{{\xi }^{3/2}}d\xi$ for a constant $c>0$. Then:

Step 4: ${\sum }_{i,j}\cos (\xi (x_i-x_j))=|{\sum }_i{e}^{i\xi x_i}{|}^2\ge 0$ and ${\sum }_{i,j}\cos (\xi (x_i+x_j))=({\sum }_i\cos (\xi x_i){)}^2-({\sum }_i\sin (\xi x_i){)}^2$.

The key comparison: ${\sum }_{i,j}(1-\cos \xi (x_i-x_j))=n^2-|S(\xi ){|}^2$ where $S(\xi )=\sum e^{i\xi x_k}$, and ${\sum }_{i,j}(1-\cos \xi (x_i+x_j))=n^2-(\text{Re}S{)}^2+(\text{Im}S{)}^2$.

Since $|S{|}^2=(\text{Re}S{)}^2+(\text{Im}S{)}^2\ge (\text{Re}S{)}^2-(\text{Im}S{)}^2$... we need $n^2-|S{|}^2\le n^2-(\text{Re}S{)}^2+(\text{Im}S{)}^2$, i.e., $(\text{Im}S{)}^2+(\text{Im}S{)}^2\ge 0$, i.e., $2(\text{Im}S{)}^2\ge 0$. This is always true!

Conclusion: The inequality holds with equality iff all $\text{Im}S(\xi )=0$ for all $\xi$, which happens when $x_i=0$ for all $i$, or more generally when the multiset $\{x_i\}$ is symmetric about 0. $■$