TITLE: Inequality of absolute values of complex sums
QUESTION [6 upvotes]: 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?

REPLY [6 votes]: 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
$$
RHS =
\left| \sum_k \frac{c_k}{x_k-ix} \right|
\ge \mathrm{Im\,} \sum_k \frac{c_k}{x_k-ix}
= \sum_k \mathrm{Im\,} \frac{c_k(x_k+ix)}{(x_k-ix)(x_k+ix)} = \\
= \sum_k \frac{c_kx}{|x_k-ix|^2}
= \sum_k \left|\frac{c_kx}{(x_k-ix)^2}\right|
\ge \left| \sum_k \frac{c_kx}{(x_k-ix)^2}\right|
= LHS.
$$