Given a finite set of distinct primitive Dirichlet characters, $\chi_1, \dots, \chi_r$, is it known that the product of the L-functions, $$L(s):=\prod_{i=1}^r L(s,\chi_i),$$ has a simple zero? It's conjectured that all of the zeros of the $L(s,\chi_i)$ are distinct and simple, but I don't know what is known unconditionally except in the case that $r\leq 2$.

It's known that each $L(s,\chi)$ has infinitely many simple zeros (in fact, a positive proportion of its zeros are simple and lie on the half-line), which immediately answers the question in the case that $r=1$. If $r=2$, the answer to my question seems to be provided by work of Conrey, Ghosh, and Gonek (Simple zeros of the zeta function of a quadratic number field, I. Invent. Math., MR0860683). They prove that the Dedekind zeta function associated to a quadratic field has $\gg T^{6/11}$ simple zeros with imaginary part up to $T$, all arising from $\zeta(s)$. It appears that their method can be adapted to consider the product of any two Dirichlet L-functions, and this is confirmed by a statement of Bombieri and Perelli (Distinct zeros of L-functions. Acta Arith., MR1611193), who additionally write that $r=2$ is the limit of the Conrey-Ghosh-Gonek method.

I have not been able to find any work which applies to my question in the case that $r\geq 3$. The paper of Bombieri and Perelli referenced above discusses counting distinct zeros of more general L-functions, but it is not obvious to me how to isolate the simple zeros in their argument.

I also don't know if the fact that I'm only looking for a single simple zero of $L(s)$ saves me anything. That is, I don't know of techniques that detect the existence of such a zero without proving that there are an infinite number. Nevertheless, this could prove to be useful, since it seems entirely possible that showing that $L(s)$ has infinitely many simple zeros when $r\geq 3$ could be quite difficult.