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I have found that $\left|\frac{2(\zeta'(x))^2-\zeta''(x)\zeta(x)}{\zeta^3(x)}\right|\leq \frac{2}{(x-\frac{1}{2})^2}$ for all real $x$ such that $x>1$ seems to be true. I have plotted the inequality and got this inequality holding. Can someone help me to prove it? here is the plot: enter image description here

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2 Answers 2

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$\newcommand\z\zeta\newcommand{\ga}{\gamma}\newcommand{\de}{\delta}$Yes, the inequality in question is true.

Indeed, this inequality is $$\frac{|g(x)|}{\z(x)^3}\le\frac2{(x-1/2)^2}$$ for $x>1$, where \begin{equation} g:=2{\z'}^2-\z''\z, \end{equation} It was shown in the previous answer that $g$ is increasing (on $(1,\infty)$) to $g(\infty-)=0$. So, $g<0$. So, the inequality in question can be rewritten as $$h(x):=\frac{\z''(x)\z(x)-2{\z'(x)}^2}{\z(x)^3}\le\frac2{(x-1/2)^2}; \tag{10}\label{10}$$ here and in what follows, $x>1$ by default.

In the same mentioned answer, it was shown that \begin{equation} \z''(x)\le z_2(x):=\frac{\ln^2 2}{2^x}+\frac{19}{10}\frac1{(5/2)^x} \end{equation} for $x\ge7/2$. Also, $z_2(x)\le\frac2{(x-1/2)^2}$ and $h\le\z''/\z^2\le\z''\le z_2$. So, \eqref{10} follows for $x\ge7/2=\frac{35}{10}$.

Also, similarly to (20) in the same mentioned answer, here we have $$h(x)\le \dfrac{\Big(\dfrac1{x-1}+\ga+\tilde r_0(x)\Big) \Big(\dfrac2{(x-1)^3}+r_2(x)\Big) -2\Big(\dfrac2{(x-1)^2}-r_1(x)\Big)^2} {\Big(\dfrac1{x-1}+\ga+\tilde r_0(x)\Big)^3} \\ <\frac2{(x-1/2)^2} \tag{30}\label{30}$$ for $x\in(1,\frac{13}{10}]$.

It remains to prove \eqref{10} for $x\in[\frac{13}{10},\frac{35}{10})$, which can be done by the interval method, again as in the same mentioned answer. $\quad\Box$

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  • $\begingroup$ Thank you very much ;) $\endgroup$
    – Haidara
    Commented Nov 10 at 17:00
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Here is a computer-based proof with Mathematica 14.1 on Windows.

NMaximize[{RealAbs[(2*D[Zeta[x], x]^2 - D[Zeta[x], {x, 2}]*Zeta[x])/
 Zeta[x]^3]*(x - 1/2)^2/2, x >= 1}, x]

{0.398781, {x -> 2.41209}}

Addition. In order to dispel doubts formulated in the comments, let us consider that optimization problem in Maple 2024, making use of a global optimizer DirectSearch

DirectSearch:-GlobalOptima(1/2*abs(2*Zeta(1, x)^2 - Zeta(2, x)*Zeta(x))*(x - 1/2)^2/Zeta(x)^3, {1 <= x}, maximize);

[0.398780684065710, [x = 2.41209275038767], 170]

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    $\begingroup$ I doubt that this can be considered a `proof'. $\endgroup$ Commented Nov 4 at 10:54
  • $\begingroup$ @PeterMueller: Can you kindly ground your doubt? In other case this is empty talk. $\endgroup$
    – user64494
    Commented Nov 4 at 12:24
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    $\begingroup$ There are many reasons why this isn't a proof, the most compelling one is in Mathematica's own documentation of the function Nmaximize: `For nonlinear functions, NMaximize may sometimes find only a local maximum for certain methods'. $\endgroup$ Commented Nov 4 at 12:49
  • $\begingroup$ @PeterMueller: Thank you for your valuable comment. I'll think about it. $\endgroup$
    – user64494
    Commented Nov 4 at 13:11
  • $\begingroup$ @PeterMueller yes, I didn't consider it as a proof but it is a nice and helpful method to check such conjectures. $\endgroup$
    – Haidara
    Commented Nov 4 at 14:28

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