A Surprising Proof of an Inequality

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

$$\sum _{i=1}^n\sum _{j=1}^n{|x_i-x_j|}^r\le \sum _{i=1}^n\sum _{j=1}^n{|x_i+x_j|}^r$$

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

$$\mathcal{I}(\alpha )={\int }_0^{\mathrm{\infty }}\frac{1-\cos \left(\alpha x\right)}{x^{r+1}}\text{d}x$$

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

$$\begin{array}{rl}|a+b{|}^r-|a-b{|}^r& =\frac{1}{\mathcal{I}(1)}{\int }_0^{\mathrm{\infty }}\frac{\cos ((a-b)x)-\cos ((a+b)x)}{x^{r+1}}\text{d}x\\ & =\frac{1}{\mathcal{I}(1)}{\int }_0^{\mathrm{\infty }}\frac{2\sin (ax)\sin (bx)}{x^{r+1}}\text{d}x\end{array}$$

and hence

$$\sum _{i=1}^n\sum _{j=1}^n{|x_i-x_j|}^r-\sum _{i=1}^n\sum _{j=1}^n{|x_i+x_j|}^r=\frac{2}{\mathcal{I}(1)}{\int }_0^{\mathrm{\infty }}\frac{{(\sum _{i=1}^n\sin (x_{i}x))}^2}{x^{r+1}}\text{d}x\ge 0$$

hence completing the proof.