Let $A$ be your ring and $X=\mathrm{Spec} A$. The minimal primes of $A$ correspond to the irreducible components of $X$. An element of $f\in A$ induces a function $\widehat f:X\to \coprod_{\mathfrak p\in \mathrm{Spec} A}\kappa(\mathfrak p)$ and this function vanishes everywhere if and only if $f\in\mathfrak p$ for all $\mathfrak p\in \mathrm{Spec} A$, that is, when it is nilpotent. For an $f\in A$ contained in a minimal prime the induced function $\widehat f$ vanishes on the irreducible component corresponding to the minimal prime containing $f$.
If $X$ has a single irreducible component, then the functions induced by the elements of $A$ in the corresponding single prime ideal are vanishing everywhere hence they are nilpotent (in particular zero-divisors).
If $X$ has more than one irreducible component, then taking one element in each minimal prime, their product will produce a function that is vanishing everywhere and hence nilpotent. It is easy to see that the product of any proper subset of these functions will not vanish on at least one irreducible component and hence it is not nilpotent, so the following claim implies that all of these are zero-divisors.
Claim Let $f_1,\dots,f_t$ be such that $f_1\cdots f_t$ is nilpotent, but $f_2\cdots f_t$ is not nilpotent, then $f_1$ is a zero-divisor.
Proof Let $(n_1,\dots,n_t)\in\mathbb N^t$ be a smallest $t$-uple (in the lexicographical order) such that $f_1^{n_1}\cdots f_t^{n_t}=0$. Then $f_1\cdot (f_1^{n_1-1}\cdots f_t^{n_t})=0$, but $f_1^{n_1-1}\cdots f_t^{n_t}\neq 0$. $\square$
Comment Apparently this is essentially the same proof as the one Graham included in the comments, but I can't let go any chance of giving a geometric proof of an algebra question. Also, it clarifies the unclear step pointed out by Georges. It appears longer, but only because i) setup ii) more details. Also, let me add that the length is the main reason I did not include it as a comment. In any case algebra=geometry. :)