Zeta bounds in classical zero free region

Under countable choice, choose $c_1$ smaller than the constant in the horizontal comparison. This gives the stated derivative bound throughout the high-height region.

At $s_1=1+1\mathrm{/}L+it$, $L=\log (\mathrm{\mid }t\mathrm{\mid }+2)$, the Euler logarithm satisfies $\mathrm{\mid }\log \zeta (s_1)\mathrm{\mid }\le \log \zeta (1+1\mathrm{/}L)\le \log (1+L)$, by comparing the positive real zeta series to its integral. Integrate ${\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta$ from $s_1$ horizontally to $s$ for $1-c_2\mathrm{/}L\le \sigma \le 1+1\mathrm{/}L$. The length is $O(1\mathrm{/}L)$ and the integrand is $O(L)$, so the change in the continued logarithm is $O(1)$. Exponentiating its negative real part gives $\mathrm{\mid }1\mathrm{/}\zeta (s)\mathrm{\mid }=O(L)$. For larger sigma the Euler logarithm already gives that bound.

On the compact low-height portion choose $c_2<c_1$ sufficiently small that $h(s)=(s-1)\zeta (s)$ is holomorphic and nonvanishing on a neighborhood, including $h(1)=1$. Then $h^{\mathrm{\prime }}\mathrm{/}h$ and $1\mathrm{/}h$ are bounded there. The identities ${\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta +1\mathrm{/}(s-1)=h^{\mathrm{\prime }}\mathrm{/}h$ and $1\mathrm{/}\zeta =(s-1)\mathrm{/}h$ prove both low-height estimates and their removable interpretations.