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Algebra Difficulty 4.6 AIME Prove it Ireland

Suppose x1,x2,…,xnx_1, x_2, \dots, x_n are complex numbers. Prove that
∑i,j=1n∣xi−xj∣2≤∑i,j=1n∣xi+xj∣2, \sum_{i,j=1}^{n} |x_i - x_j|^2 \le \sum_{i,j=1}^{n} |x_i + x_j|^2,
with equality iff x1+x2+⋯+xn=0x_1 + x_2 + \dots + x_n = 0.

Solution

Note that ∣a+b∣2−∣a−b∣2=4Re⁡(aˉb)|a+b|^2 - |a-b|^2 = 4 \operatorname{Re}(\bar{a}b) for any complex numbers a,ba, b. Hence,
∑i,j=1n∣xi+xj∣2−∑i,j=1n∣xi−xj∣2=4∑i,j=1nRe⁡(xˉixj)=4Re⁡∑i,j=1nxˉixj=4Re⁡∑i=1nxˉi∑j=1nxj=4∣∑i=1nxi∣2≥0, \begin{align*} \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} \operatorname{Re} (\bar{x}_i x_j) = 4 \operatorname{Re} \sum_{i,j=1}^{n} \bar{x}_i x_j \\ &= 4 \operatorname{Re} \sum_{i=1}^{n} \bar{x}_i \sum_{j=1}^{n} x_j = 4 \left| \sum_{i=1}^{n} x_i \right|^2 \geq 0, \end{align*}
and equality holds iff ∑i=1nxi=0\sum_{i=1}^{n} x_i = 0.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.