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It's easy to prove, for example, that $|\zeta(2 + it)| > 2 - \frac{\pi^2}{6}$. However, there is some $\sigma > 1$ for which $\zeta ( \sigma ) = 2$, and it is more difficult to obtain a lower bound on $|\zeta (\sigma' + it) |$ for $1<\sigma' \leq \sigma$. Can you provide a non-trivial lower bound?

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Claim: $$ |\zeta(s)| >\frac{\pi^2}{15}(\sigma-1). $$ Proof: We compare Euler products to see $ 1/|\zeta(s)|<\zeta(\sigma)/\zeta(2\sigma) $, so $$ |\zeta(s)|>\frac{\zeta(2\sigma)}{\zeta(\sigma)}>\frac{\pi^2}{15}(\sigma-1), $$ since the linear lower bound is tight at $\sigma=1$ and $\sigma=2$ and the function is concave down.

For $\sigma=1$, there's no known lower bound that's independent of $t$ (and thus constant.) Titchmarsh Theory of the Riemann Zeta Function proves that $1/\zeta(1+i t)=O(\log(t)^7)$ so $$ |\zeta(1+i t)|> C\log(t)^{-7}, $$ for some $C$ which can be made explicit by carefully reading his proof.

There are some more modern results - Kevin Ford has a couple of papers (with Riemann Zeta Function in the title.)

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  • $\begingroup$ Is it possible to do better for $Re(s) = 1$? $\endgroup$ – Craig Oct 20 '11 at 19:53

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