There is an almost invisible, but significant difference between the question below and that recently answered by Boris Bukh.
Given a probability measure $\mu$ supported on a finite set $S\subset{\mathbb R}^2$, define $$ f(z):=\max\left\{\mu(x)\colon \frac{x+y}2=z,\ x,y\in S \right\}, \ z\in{\mathbb R}^2. $$ Now let $\bar\mu$ be the uniform probability measure supported on the same set $S$, and define $$ \bar f(z):=\max\left\{\bar\mu(x)\colon \frac{x+y}2=z,\ x,y\in S \right\},\ z\in{\mathbb R}^2. $$
Is it true that for any choice of the measure $\mu$, the total mass of $f$ is at least as large as the total mass of $\bar f$?
(Notice that $\bar f$ is actually a scaled indicator function of the set $\{(x+y)/2\colon x,y\in S\}$, and that the total mass of $\bar f$ is actually the doubling coefficient $|S+S|/|S|$.)