= Reciprocal zeta bounds near the line one
{title2=$|\zeta(\sigma+it)^{-1}|\ll(\sigma-1)^{-3/4}\log^{1/2}(2+|t|)$}
For $1<\sigma\le2$, the <three-four-one inequality for Euler products> gives $|1/\zeta(\sigma+it)|\le\zeta(\sigma)^{3/4}|\zeta(\sigma+2it)|^{1/4}$. For $|t|\ge1$, the <fractional-part continuation formula for the Riemann zeta function>, truncated at $H\asymp2+|t|$, gives $\zeta^{(j)}(\sigma+it)=O_j(\log^{j+1}(2+|t|))$. Thus the reciprocal bound even holds with exponent $1/4$ instead of $1/2$ in this range. On bounded-height compact sets, <three-four-one product proof of zeta boundary nonvanishing> and removal of the reciprocal's zero at the pole give uniform bounds. Differentiating $1/\zeta$ repeatedly yields
$$
\left|\left(\frac1\zeta\right)^{(k)}(\sigma+it)\right|
\ll_k(\sigma-1)^{-3(k+1)/4}\log^{3k+1}(2+|t|).
$$
The range restriction matters: $1/\zeta(\sigma)\to1$ as $\sigma\to\infty$, so the first displayed bound cannot hold uniformly for all $\sigma>1$. A global version replaces $(\sigma-1)^{-3/4}$ with $1+(\sigma-1)^{-3/4}$.
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