Yes your bound is optimal and here's how to attain it. Fix a linear ordering $L$ of $\Omega$. Pick a number $x$ in [0,1] under the uniform measure.

Let your random variable realizing $\mu$ be as follows. Your outcome in $\Omega$ is $a$ such that the $\mu$ probability of an outcome $L$-below $a$ is $x$.

Similarly for $\nu$ with $x+1/2$ mod 1:


Let your random variable realizing $\nu$ be as follows. Your outcome in $\Omega$ is the $L$-greatest $b$ such that the $\nu$ probability of an outcome $L$-below $b$ is $\le x+1/2$ mod 1.

This makes $\pi(\Delta)=0$ unless there is a single outcome whose $\mu+\nu>1$, by the way.