Given a probability measure $\mu$ supported on a finite set $S\subset{\mathbb R}^2$, define 
 $$ f(z):=\max\left\{\frac{\mu(x)+\mu(y)}2\colon \frac{x+y}2=z,\ x,y\in S \right\},
 \ z\in{\mathbb R}^2. $$
Now let $\bar\mu$ be the *uniform* probability measure supported on the same set $S$, and define
 $$ \bar f(z):=\max\left\{\frac{\bar\mu(x)+\bar\mu(y)}2\colon \frac{x+y}2=z,\ x,y\in S \right\},\ z\in{\mathbb R}^2. $$
(Thus $\bar f$ is actually a scaled indicator function of the set $\{(x+y)/2\colon x,y\in S\}$.)



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> 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 $\bar f$?
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As a numerical illustration: if the points of $S$ are in a general position (meaning that $x+y=x'+y'$ with $x,y,x',y'\in S$ is only possible for $\{x,y\}=\{x',y'\}$), then the total masses of both $f$ and $\bar f$ are equal to $\frac12(|S|+1)$.



Smoothening a probability measure
Not true. Let $S=\{1,2,\dotsc,n\}\cup\{2n\}$. The total mass of $\bar{f}$ is $\approx 3n\cdot (1/n)=3$. Let $\mu$ be the uniform measure on $\{1,2,\dotsc,n\}$. For this choice of $\mu$, the total mass of $f$ is $\approx 2n\cdot (1/n)+n\cdot(1/2n)=2.5$. Here $2n$ counts the elements $\{2,3,\dotsc,2n\}$ and $n$ counts the elements from $2n+1$ to $3n$. (If you do not like that $\mu(2n)=0$, then make $\mu(2n)=\varepsilon$ very small.)


The opposite inequality also fails, as is witnessed by making $\mu(2n)$ in this example big, not small.


