There is an almost invisible, but significant difference between the question below and [that](https://mathoverflow.net/questions/215148/smoothening-a-probability-measure) 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* $|S+S|/|S|$.)



Smoothening a measure, II
This is also false.


Let $S=A\cup B$ where $A$ is $\{0,1,\dotsc,n-1\}$ and $B$ is a subset of $\{n,n+1,\dotsc,2n\}$ of size $m\leq 4\sqrt{n}$ such that $B+B$ contains all the integers from $2n$ to $4n$. (We can take $B$ to be a union of all multiples of $\sqrt{n}$ and two suitable arithmetic progressions of step $1$.) Note that the mass of $\bar{f}$ is $\approx 4n\cdot (1/n)=4$. Consider the probability distribution that is uniform on $A$, and assigns zero (or nearly zero) weight to $B$. For such a distribution, we compute the total weight of $f$ to be $\approx 3n\cdot(1/n)+n\cdot 0=3$.


I am not sure what exactly you are trying to do, but it is extremely unlikely that an inequality of the approximate form that you seek exists. The reason is that it must behave well under disjoint unions, tensor products, and be stable under local perturbations. That is a lot to ask!


