IRL_ABooklet

Note that $|a+b{|}^2-|a-b{|}^2=4\mathrm{Re}(\bar{a}b)$ for any complex numbers $a,b$. Hence, $$\begin{array}{rl}\sum _{i,j=1}^n|x_i+x_j{|}^2-\sum _{i,j=1}^n|x_i-x_j{|}^2& =4\sum _{i,j=1}^n\mathrm{Re}({\bar{x}}_i{x}_j)=4\mathrm{Re}\sum _{i,j=1}^n{\bar{x}}_i{x}_j\\ & =4\mathrm{Re}\sum _{i=1}^n{\bar{x}}_i\sum _{j=1}^n{x}_j=4{\left|\sum _{i=1}^n{x}_i\right|}^2\ge 0,\end{array}$$ and equality holds iff ${\sum }_{i=1}^n{x}_i=0$.