Let $\mathcal{M}$ be a compact Riemannian manifold with geodesic distance function $d$ and $\Omega$ its volume measure.
Pick some $A,B\subseteq\mathcal{M}$ such that $\Omega(A)\ll\Omega(B)$, but: (1) $A$ is not contained in $B$ (otherwise what I want to ask is trivial) and (2) $A$ is not contained in $B$ after an isometry of $\mathcal{M}$.
Let $f:[0,\infty)\to[0,\infty)$ be some nice smooth monotone decreasing function.
How to prove (is it even true??) that $$ \int_{x,y\in A}f(d(x,y))\mathrm{d}\Omega(x)\mathrm{d}\Omega(y) \leq \int_{x,y\in B}f(d(x,y))\mathrm{d}\Omega(x)\mathrm{d}\Omega(y)\,? $$
The statement is clear in case $f=1$ because then we simply have $\Omega(A)^2 \ll \Omega(B)^2$ which is true as $\cdot^2$ is monotone increasing. However, when $f$ is not equal to a constant and there is no isometry bringing $A$ into $B$, I am not sure how to approach this, though it seems intuitive that it must be true, or at least, can anyone think of a counter-example?
Let me write $$V(A)=\int_{x,y\in A}f(d(x,y))\mathrm{d}\Omega(x)\mathrm{d}\Omega(y).$$ Note that since $f$ is decreasing, $f(r)\leq f(0)$ for all $r$ so $V(A)\leq f(0)\Omega(A)^2.$ In particular, this goes to $0$ as $\Omega(A)\to 0$, so for any fixed $B$ of positive measure, $V(A)\leq V(B)$ for all $A$ of sufficiently small measure.
(Note that unless $f$ is identically $0$, $V(B)>0$ for any $B$ of positive measure, since $B$ must have positive measure in some small ball in which $f(d(\cdot,\cdot))$ is always close to $f(0)$. In fact, using compactness of $\mathcal{M}$ to cover it with finitely many (say, $n$) small balls, $B$ must have measure at least $\Omega(B)/n$ in one of those balls, and so we get a lower bound on $V(B)$ depending only on $\Omega(B)$ (and $f$). So, the corresponding upper bound on $\Omega(A)$ needed to guarantee $V(A)\leq V(B)$ depends only on $\Omega(B)$ (and $f$), not on $B$ itself.)