<s>[INST] Prove that $\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}> \frac{3\sqrt{3}}{2}(\zeta(2s)-1),\space s>1$ What ways would you propose for getting the inequality below? 
$$\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}> \frac{3\sqrt{3}}{2}(\zeta(2s)-1),\space s>1$$
The left side may be written as 
$$\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}>\zeta(s)-1$$
but how do we prove then that 
$$\zeta(s)-1> \frac{3\sqrt{3}}{2}(\zeta(2s)-1)$$
?
Can we do it without using integrals?
 [/INST] From $n^s\ge3^s>2^s+1$ when $n\ge3$, we have
\begin{align}
&(\zeta(s)-1)-(2^s+1)(\zeta(2s)-1)\\&=\sum_{n=2}^\infty\frac1{n^s}\left(1-\frac{2^s+1}{n^s}\right)\\&=\sum_{k=1}^\infty\frac{1}{2^{ks}}\left(1-\frac{2^s+1}{2^{ks}}\right)+\sum_{\substack{n\ge3\\n\ne2^k}}\frac{1}{n^{s}}\left(1-\frac{2^s+1}{n^s}\right)\\&>\sum_{k=1}^\infty\frac{1}{2^{ks}}\left(1-\frac{2^s+1}{2^{ks}}\right)+\sum_{\substack{n\ge3\\n\ne2^k}}\frac{1}{n^{s}}\left(1-\frac{n^s}{n^s}\right)\\&=\sum_{k=1}^\infty\frac{1}{2^{ks}}\left(1-\frac{2^s+1}{2^{ks}}\right)\\&=\sum_{k=1}^\infty\left(\frac{1}{2^{ks}}-\frac{1}{2^{(2k-1)s}}-\frac{1}{2^{2ks}}\right)=0
\end{align}
So$$\frac{\zeta(s)-1}{\zeta(2s)-1}>2^s+1>3>\frac{3\sqrt{3}}{2}$$and we are done.
We can also prove $\zeta(s)>\zeta(2s)^3$ using $\displaystyle\zeta(s)=\prod_p\left(1-\frac1{p^s}\right)^{-1}$ and finish by $\displaystyle\frac{\zeta(s)-1}{\zeta(2s)-1}>\zeta(2s)^2+\zeta(2s)+1>3$ but I think the above way is more cleaner. To add some more info on this direction, Mathematica plot suggests $\zeta(s)>\zeta(2s)^5$ hold.

Proving the first proposed inequality is much more easier... We just need to prove $$\frac{x}{x^2-1}>\frac{3\sqrt{3}}{2}\frac{1}{x^2}$$ for $x>2$. $\frac{x^3}{x^2-1}$ has minimum at $x=\sqrt{3}$ so $\frac{x^3}{x^2-1}>\frac{3\sqrt{3}}{2}$. Perhaps that's where the constant $\frac{3\sqrt{3}}{2}$ came from?
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