IMO 2021, Problem 2.
Let $ n $ be an integer $ \ge 2$ and $x_1,x_2,...,x_n $ be $ n$ reals. prove that
$$\sum_{i=1}^n\sum_{j=1}^n\sqrt{|x_i-x_j|}\le \sum_{i=1}^n\sum_{j=1}^n\sqrt{|x_i+x_j|}$$
I wrote the left sum as $$2\sum_{i=2}^n\sum_{j=1}^{i-1}\sqrt{|x_i-x_j|}$$ and the right one as $$2\sum_{i=2}^n\sum_{j=1}^{i-1}\sqrt{|x_i+x_j|}+\sum_{i=1}^n\sqrt{2|x_i|}$$
but, it seems this is not a good starting point.
This is certainly not an answer expected in IMO, but the proof works for far more general cases, see this posting, for instance.
Note that
$$ \frac{1}{c} \int_{0}^{\infty} \frac{1-\cos (as)}{s^{3/2}} \, \mathrm{d}s = |a|^{1/2} \quad \text{for any} \quad a \in \mathbb{R}, $$
where $c = \int_{0}^{\infty} \frac{1-\cos t}{t^{3/2}} \, \mathrm{d}t \in (0, \infty)$. This is easily proved by substituting $t = |a|s$. Then
\begin{align*} &\sum_{i,j} |x_i + x_j|^{1/2} - \sum_{i,j} |x_i - x_j|^{1/2} \\ &= \frac{1}{c} \int_{0}^{\infty} \frac{1}{s^{3/2}}\biggl( \sum_{i,j} \cos((x_i- x_j)s) - \cos((x_i + x_j)s) \biggr) \, \mathrm{d}s \\ &= \frac{1}{c} \int_{0}^{\infty} \frac{1}{s^{3/2}}\biggl( \sum_{i,j} 2\sin(x_i s)\sin(x_j s) \biggr) \, \mathrm{d}s \\ &= \frac{1}{c} \int_{0}^{\infty} \frac{2}{s^{3/2}}\biggl( \sum_{i} \sin(x_i s) \biggr)^2 \, \mathrm{d}s \\ &\geq 0. \end{align*}
Of course, it would be fun to have a more elementary solution.
