We have an exact formula \begin{align*} \frac{1}{\zeta'(\rho)} &= \lim_{s \to \rho} \frac{s-\rho}{\zeta(s)} \\ &= \lim_{s \to \rho} \frac{(s-\rho) (s-1) \Gamma(1+s/2) \pi^{-s/2}}{\xi(s)} \\ &= (\rho-1) \Gamma(1+\rho/2) \pi^{-\rho/2} \lim_{s \to \rho} \frac{s-\rho}{\frac{1}{2} e^{Bs} \prod_{\rho'} (1-\frac{s}{\rho'}) e^{-s/\rho'}} \\ &= -\frac{ 2 e^{1-B\rho} \rho(\rho-1) \Gamma(1+\rho/2) \pi^{-\rho/2}}{\prod_{\rho' \neq \rho} (1-\frac{\rho}{\rho'}) e^{-\rho/\rho'}} \end{align*} where $B = -0.0230957\dots$ is the constant in Theorem 10.12 of Montgomery-Vaughan. All of the factors in the above formula are well understood except for the terms $1-\frac{\rho}{\rho'}$ for nearby zeroes $\rho' = \rho+O(1)$, which are proportional in magnitude to the distances from $\rho$ to the nearby zeroes $\rho'$. So the problem of upper bounding $1/\zeta'(\rho)$ is more or less equivalent to that of lower bounding the distance $|\rho-\rho'|$ to the nearest zero $\rho'$ (or more precisely the product of the distances to those zeroes $\rho'$ within $O(1)$ of $\rho$). An assumption of simple zeroes merely says that this distance is positive, but a more quantitative version of this hypothesis would be needed to get any quantitative upper bound.

We have an exact formula \begin{align*} \frac{1}{\zeta'(\rho)} &= \lim_{s \to \rho} \frac{s-\rho}{\zeta(s)} \\ &= \lim_{s \to \rho} \frac{(s-\rho) (s-1) \Gamma(1+s/2) \pi^{-s/2}}{\xi(s)} \\ &= (\rho-1) \Gamma(1+\rho/2) \pi^{-\rho/2} \lim_{s \to \rho} \frac{s-\rho}{\frac{1}{2} e^{Bs} \prod_{\rho'} (1-\frac{s}{\rho'}) e^{s/\rho'}} \\ &= -\frac{ 2 e^{1-B\rho} \rho(\rho-1) \Gamma(1+\rho/2) \pi^{-\rho/2}}{\prod_{\rho' \neq \rho} (1-\frac{\rho}{\rho'}) e^{\rho/\rho'}} \end{align*} where $B = -0.0230957\dots$ is the constant in Theorem 10.12 of Montgomery-Vaughan. All of the factors in the above formula are well understood except for the terms $1-\frac{\rho}{\rho'}$ for nearby zeroes $\rho' = \rho+O(1)$, which are proportional in magnitude to the distances from $\rho$ to the nearby zeroes $\rho'$. So the problem of upper bounding $1/\zeta'(\rho)$ is more or less equivalent to that of lower bounding the distance $|\rho-\rho'|$ to the nearest zero $\rho'$ (or more precisely the product of the distances to those zeroes $\rho'$ within $O(1)$ of $\rho$). An assumption of simple zeroes merely says that this distance is positive, but a more quantitative version of this hypothesis would be needed to get any quantitative upper bound.

We have an exact formula \begin{align*} \frac{1}{\zeta'(\rho)} &= \lim_{s \to \rho} \frac{s-\rho}{\zeta(s)} \\ &= \lim_{s \to \rho} \frac{(s-\rho) (s-1) \Gamma(1+s/2) \pi^{-s/2}}{\xi(s)} \\ &= (\rho-1) \Gamma(1+\rho/2) \pi^{-\rho/2} \lim_{s \to \rho} \frac{s-\rho}{\frac{1}{2} e^{Bs} \prod_{\rho'} (1-\frac{s}{\rho'}) e^{-s/\rho'}} \\ &= -\frac{ 2 e^{1-B\rho} \rho(\rho-1) \Gamma(1+\rho/2) \pi^{-\rho/2}}{\prod_{\rho' \neq \rho} (1-\frac{\rho}{\rho'}) e^{-\rho/\rho'}} \end{align*} where $B$ is the constant in Theorem 10.12 of Montgomery-Vaughan. All of the factors in the above formula are well understood except for the terms $1-\frac{\rho}{\rho'}$ for nearby zeroes $\rho' = \rho+O(1)$, which are proportional in magnitude to the distances from $\rho$ to the nearby zeroes $\rho'$. So the problem of upper bounding $1/\zeta'(\rho)$ is more or less equivalent to that of lower bounding the distance $|\rho-\rho'|$ to the nearest zero $\rho'$ (or more precisely the product of the distances to those zeroes $\rho'$ within $O(1)$ of $\rho$). An assumption of simple zeroes merely says that this distance is positive, but a more quantitative version of this hypothesis would be needed to get any quantitative upper bound.

We have an exact formula \begin{align*} \frac{1}{\zeta'(\rho)} &= \lim_{s \to \rho} \frac{s-\rho}{\zeta(s)} \\ &= \lim_{s \to \rho} \frac{(s-\rho) (s-1) \Gamma(1+s/2) \pi^{-s/2}}{\xi(s)} \\ &= (\rho-1) \Gamma(1+\rho/2) \pi^{-\rho/2} \lim_{s \to \rho} \frac{s-\rho}{\frac{1}{2} e^{Bs} \prod_{\rho'} (1-\frac{s}{\rho'}) e^{-s/\rho'}} \\ &= -\frac{ 2 e^{1-B\rho} \rho(\rho-1) \Gamma(1+\rho/2) \pi^{-\rho/2}}{\prod_{\rho' \neq \rho} (1-\frac{\rho}{\rho'}) e^{-\rho/\rho'}} \end{align*} where $B = -0.0230957\dots$ is the constant in Theorem 10.12 of Montgomery-Vaughan. All of the factors in the above formula are well understood except for the terms $1-\frac{\rho}{\rho'}$ for nearby zeroes $\rho' = \rho+O(1)$, which are proportional in magnitude to the distances from $\rho$ to the nearby zeroes $\rho'$. So the problem of upper bounding $1/\zeta'(\rho)$ is more or less equivalent to that of lower bounding the distance $|\rho-\rho'|$ to the nearest zero $\rho'$ (or more precisely the product of the distances to those zeroes $\rho'$ within $O(1)$ of $\rho$). An assumption of simple zeroes merely says that this distance is positive, but a more quantitative version of this hypothesis would be needed to get any quantitative upper bound.