I botched up a proof, but I salvaged useful bits and bobs from it.

We prove \(\zeta^2(s) < \zeta(s+1)\zeta(s-1)\) by recalling the Dirichlet series of \(\phi\), the Euler totient function, that is,

\[\sum_{n=1}^{\infty} \frac{\phi(n)}{n^s} = \frac{\zeta(s-1)}{\zeta(s)}\]

Now we know that, for all \(n \geq 2\),

\[\frac{\phi(n)}{n^{s+1}} < \frac{\phi(n)}{n^s} \longrightarrow \sum_{n=2}^{\infty} \frac{\phi(n)}{n^{s+1}} < \sum_{n=2}^{\infty} \frac{\phi(n)}{n^s} \quad \therefore \frac{\zeta(s)}{\zeta(s-1)} < \frac{\zeta(s+1)}{\zeta(s)}\]

\[\therefore \zeta^2(s) < \zeta(s+1)\zeta(s-1)\]

Personally, I think it is interesting, as it invokes a Dirichlet series in a somewhat unexpected way. Also, this implies that \(\lbrace \frac{\zeta(n+1)}{\zeta(n)} \rbrace _{n=2}\) is monotonically increasing.

Challenge: Prove \(2\zeta(s) < \zeta(s+1)+\zeta(s-1)\), using the above result or by the trivial inequality.

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## Comments

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TopNewestJust prove that \( \zeta(s) \) concave downwards, the inequality follows.

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I just realised, Cauchy-Schwarz on \(\ell^2\) works.

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Since if \(a,b>0\), \(a^{2}+b^{2}>2ab\)

\[2\zeta (s+1)\zeta (s-1)<{ \zeta (s+1) }^{ 2 }+{ \zeta (s-1) }^{ 2 }\]

Following the note's inequality, \[2{ \zeta (s) }^{ 2 }<2\zeta (s+1)\zeta (s-1)<{ \zeta (s+1) }^{ 2 }+{ \zeta (s-1) }^{ 2 }\] \[4{ \zeta (s) }^{ 2 }<{ \zeta (s+1) }^{ 2 }+{ \zeta (s-1) }^{ 2 }+2\zeta (s+1)\zeta (s-1)\] \[2\zeta (s)<{ \zeta (s+1) }+{ \zeta (s-1) }\]

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Or that. I think that spotting that there's a way to slip in AM-GM is easier, but again, that just depends on what prior experience one has.

There is also a proof without the result of the challenge by the trivial inequality, which is kind of obvious.

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I think you meant \(\lbrace \frac{\zeta(n+1)}{\zeta(n)} \rbrace _{n\ge2}\)

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Same thing, hehe.

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Maybe I misunderstood the notation...

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