Let $\mu,\nu$ be probability measures defined on a common measure space $(\Omega,\mathcal F)$. A coupling of $\mu,\nu$ is a probability measure $\pi$ on $(\Omega^2,\mathcal F^2)$ with marginals $\mu,\nu$ respectively. Let $\Delta\subseteq \Omega^2$ denote the diagonal, $\Delta=\{(\omega,\omega)\colon \omega\in \Omega^2\}$. It is well-known (and not difficult to verify) that every coupling $\pi$ satisfies the inequality $$ \pi(\Delta)\leq 1-\|\mu-\nu\|_{TV}, $$ where $\|\mu-\nu\|_{TV}:=\sup_{A\in \mathcal F}|\mu(A)-\nu(A)|$, and it is also not difficult to construct a coupling attaining the equality (i.e. constructing a maximal coupling).

My question concerns minimal couplings: is there a non-trivial lower bound on $\pi(\Delta)$, and if so, is there a simple construction attaining this lower bound for all probability measures $\mu,\nu$?

The question is less interesting in the non-atomic case, since the independent coupling would already assign no mass to the diagonal. Also, if singletons are measurable, then there is an easy lower bound of $\pi(\{(x,x)\})\geq \bigl(\mu(\{x\})+\nu(\{x\})-1\bigr)_+$, from which one can derive a lower bound on $\pi(\Delta)$ in the completely atomic case. Is this bound optimal? Under what conditions on $\mu,\nu$ does there exist a coupling with $\pi(\Delta)=0$?