# Inequality of absolute values of complex sums

Let $c_1,\dots, c_n\geq 0$ and $x,\dots,x_n\in\mathbb R$. Then $$\left\lvert \sum_k \frac{c_k x}{(x_k-i x)^2}\right\rvert\leq \left\lvert \sum_k \frac{c_k }{x_k-i x}\right\rvert$$ seems to hold for all real $x\not=0$.

Why is this the case? It's trivial for $n=1$, but already for $n=2$ writing out these absolute values becomes rather messy. Am I overlooking something?

• What is your evidence that this "seems to hold"? – Eric Wofsey Jul 4 '16 at 5:58
• For small values of $n$ I performed a numerical maximisation of the fraction of the two quantities in Mathematica. The maximum is consistently 1 and is attained when all $x_k$ are very close to $0$. For large $n$ I generated random $c$ and $\lambda$ and performed the maximisation only in $x$. – Julian Jul 4 '16 at 6:12
• Here's a heuristic explanation for why this sounds plausible: when the $x_k$ are small compared to $x$, the arguments of the terms on the right-hand side are close together and so squaring them causes them to spread out and cancel each other out more. When the $x_k$ are large compared to $x$, there is more cancellation on the right-hand side, but then the factors $x/(x_k-ix)$ will be small and can reasonably be expected to make the left-hand side smaller despite having less cancellation. – Eric Wofsey Jul 4 '16 at 6:28

The numbers $(x,x_1,\ldots,x_n)$ can be replaced by $(-x,-x_1,\ldots,-x_n)$ without changing the statement, so we may assume $x\ge0$. Then $$RHS = \left| \sum_k \frac{c_k}{x_k-ix} \right| \ge \mathrm{Im\,} \sum_k \frac{c_k}{x_k-ix} = \sum_k \mathrm{Im\,} \frac{c_k(x_k+ix)}{(x_k-ix)(x_k+ix)} = \\ = \sum_k \frac{c_kx}{|x_k-ix|^2} = \sum_k \left|\frac{c_kx}{(x_k-ix)^2}\right| \ge \left| \sum_k \frac{c_kx}{(x_k-ix)^2}\right| = LHS.$$