We start with two disjoint compact sets A and B with positive capacities. Then, we translate B s.t. $B+rv$ is disjoint from A and B and ,more importantly, $dist(x,y)<dist(x,y+rv)$ for all $x\in A$ and $y\in B$. Does this imply

$$Cap(A\sqcup B)<Cap(A\sqcup (B+rv))?$$

We have $\leq $ because contraction maps decrease capacity (Landkoff in chapter "metric properties of capacities"). The map that sents $A\sqcup B+rv$ to $A\sqcup B$ decreases distances.

I will post as I find things.

We have $Cap(A)=[inf_{\mu(A)=1}\int_{A}\int_{A}\frac{1}{|x-y|^{d-2}}d\mu(x) d\mu(y)]^{-1}$. Call the infimum measure $\mu_{A}$ (equilibrium measure).

So the diffulty is in showing that $\mu_{A\sqcup B}\geq \mu_{A\sqcup (B+rv)}$.

Why does there exist measure $\mu'_{A\cup B+rv}$ s.t. $\int\int_{A\cup B} \frac{1}{|x-y|} d\mu_{A\cup B}>\int\int_{A\cup B+rv} \frac{1}{|x-y|} d\mu'_{A\cup B+rv}$?

If we take any measure supported on $A$ and $B+rv$, it is not obvious to me that that the above inequality is true.

If we set $\mu'_{A\cup B+rv}(A):=\mu_{A\cup B}(A)$, how can we shift the measure while keeping the integral lower or equal?

Maybe just define $\mu'_{A\cup B+rv}(B+rv):=\mu_{A\cup B}(B)$