Let $\zeta$ denote the Riemann zeta function and let $\rho$ denote one of its complex zeros. What is the best known upper bound for $\frac{1}{\zeta'(\rho)}$ assuming that all zeros are simple (SZC), **but not** assuming Riemann Hypothesis ?

Assuming both the RH and the SZC, one can mimick the proof of Theorem 15.6 of Montgomery-Vaughan's *Multiplicative Number Theory* and show that
$$
\frac{1}{\zeta'(\rho)} \ll X,\label{1}\tag{1}
$$
where $X$ is any real number $\geq |\rho| $ (actually the Montgomery-Vaughan argument seems to yield an upper bound of the form $o(X)$).

However, it looks like the bound \eqref{1} could hold assuming the SZC alone, as it appears that Montgomery-Vaughan only invoked the RH on bounding the $S(T)=\arg \zeta(\sigma + iT)$. On the RH, it is a classical fact that
$$
S(T) \ll \frac{\log T}{\log \log T}
$$
whilst $S(T) \ll \log T$ unconditionally. The unconditional bound **seems** sufficient for the purposes of showing that \eqref{1} comes from the Montgomery-Vaughan argument.