A Surprising Proof of an Inequality

Problem : Prove that for real numbers $x_1,\dots ,x_n$,

$$\sum_{i=1}^n \sum_{j=1}^n \left\lvert x_i - x_j\right\rvert^r \le \sum_{i=1}^n \sum_{j=1}^n \left\lvert x_i + x_j\right\rvert^r$$

for any $r\in [0,2]$.

Solution : The cases $r=0,2$ can be checked by hand. So, we assume $r\in (0,2)$ and begin by noting that the integral

$$\mathcal I (\alpha) = \int_{0}^\infty \frac{1-\cos \left(\alpha x\right)}{x^{r+1}}\;\text{d}x$$

is positive and finite. Also, it is easy to prove using the substitution $y=|\alpha|x$ that $\mathcal I(\alpha)=\sqrt{|\alpha|}\cdot \mathcal I(1)$.

Now, using this, we have

\begin{align*} |a+b|^r-|a-b|^r &= \frac{1}{\mathcal I(1)} \int_0^\infty \frac{\cos \left((a-b)x\right) - \cos \left((a+b)x\right)}{x^{r+1}} \;\text{d}x\\ &= \frac{1}{\mathcal I(1)} \int_0^\infty \frac{2 \sin (ax) \sin (bx)}{x^{r+1}} \;\text{d}x \end{align*}

and hence

$$\sum_{i=1}^n \sum_{j=1}^n \left\lvert x_i - x_j\right\rvert^r - \sum_{i=1}^n \sum_{j=1}^n \left\lvert x_i + x_j\right\rvert^r = \frac{2}{\mathcal I(1)} \int_0^\infty \frac{\left(\sum_{i=1}^n \sin (x_i x)\right)^2}{x^{r+1}} \;\text{d}x \ge 0$$

hence completing the proof.