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Is it true that

$$|\zeta(\frac{1}{2}+it)|< \frac{1}{4}+t^2$$ for all $t\geq 0$, where $\zeta$ denotes the Riemann zeta function ?

It is known that the left hand-side is $O(t^{0.25})$, and on the the Lindelof Hypothesis, it is actually $O(t^\epsilon)$ for any positive $\epsilon$. But since these results are only valid for large $|t|$, thyey dont seem to answer my question.

Note: I have added abs value for The LHS of inequality because it is not clear if $\zeta(\frac{1}{2}+it)$ is always real number

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  • $\begingroup$ In the paper: Hiary, Ghaith A. An explicit van der Corput estimate for ζ(1/2+it). Indag. Math. (N.S.) 27 (2016), no. 2, 524–533. It is proved that $|\zeta(1/2+it)|\le 0.63 t^{1/6}\log t$ for $t\ge 3$ and we have $|\zeta(1/2+it)|\le 1.461$ for $0\le t\le 3$. Of course the proof is more complicate than that in the answer of Carlo Beenakker. $\endgroup$ – juan Feb 26 '18 at 16:18
  • $\begingroup$ Please include a top-level tag (e.g. nt.number-theory) next time! $\endgroup$ – GH from MO Feb 26 '18 at 22:59
  • $\begingroup$ You can find further useful bounds here: iml.univ-mrs.fr/~ramare/TME-EMT/Articles/Art06.html $\endgroup$ – GH from MO Feb 26 '18 at 23:02
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The inequality at the top of page 99 in Lectures on The Riemann Zeta–Function gives $$|\zeta(\tfrac{1}{2}+it)|<9t^{1/2},\;\;t\geq 1.$$ To cover the whole the interval $t\geq 0$, just add $3/2$ to the right-hand-side of the inequality.

The inequality $|\zeta(\tfrac{1}{2}+it)|<\tfrac{1}{4}+t^2$ holds for $t> 0.8$.

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Visually,

enter image description here

where black dots are values of $\zeta\left(\frac12+it\right)$ and the red ones are at the distance $\frac14+t^2$ from the origin.

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