Let $R$ be a reduced ring. Show that $\operatorname{Ass}R$ is the set of minimal prime ideals of $R$.
I think that the first inclusion must come from using $\operatorname{Ass}R \subseteq\operatorname{Supp}R$, assuming the ideal is not minimal, and then showing some contradiction. No idea how to prove every minimal prime is associate, since the ring is not necessarily noetherian.
If $R$ is reduced and $P=\operatorname{Ann}(x)$ is an associated prime, suppose $Q$ is a prime properly contained in $P$. Then $(x)P\subseteq Q$ implies $x\in Q$. But then $x^2=0$, a contradiction. So $P$ was already minimal.
The other direction isn't clear to me. In this paper they talk about necessary and sufficient conditions for a reduced ring to have the property that all finitely generated ideals of zero divisors to have a nonzero annihilator, so presumably both cases can happen.
I see here that "weakly associated primes" are exactly the minimal primes in a reduced ring though. In that case, minimal primes are obviously weakly associated, because a minimal prime is minimal over $\operatorname{Ann}(1)=\{0\}$.