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?
 A: The Mean Value Theorem says that for some $\eta$ between $n$ and $n+1$, we have
$$
n^{1-{\large s}}-(n+1)^{1-{\large s}}=(1-s)\eta^{-{\large s}}\tag{1}
$$
Thus, for $s\gt0$,
$$
\frac{n^{1-{\large s}}-(n+1)^{1-{\large s}}}{1-s}
\le n^{-{\large s}}
\le\frac{(n-1)^{1-{\large s}}-n^{1-{\large s}}}{1-s}\tag{2}
$$
We easily have the upper bound
$$
\begin{align}
\zeta(s)-1
&=2^{-{\large s}}+\sum_{n=3}^\infty n^{-{\large s}}\\
&\le2^{-{\large s}}+\sum_{n=3}^\infty\frac{(n-1)^{1-{\large s}}-n^{1-{\large s}}}{1-s}\\[4pt]
&=\frac{s+1}{s-1}2^{-{\large s}}\tag{3}
\end{align}
$$
Furthermore, we have the lower bound
$$
\begin{align}
\zeta(s)-1
&=2^{-{\large s}}+\sum_{n=3}^\infty n^{-{\large s}}\\
&\ge2^{-{\large s}}+\sum_{n=3}^\infty\frac{n^{1-{\large s}}-(n+1)^{1-{\large s}}}{1-s}\\
&=2^{-{\large s}}+\frac3{s-1}3^{-{\large s}}\tag{4}
\end{align}
$$
Therefore, using $(1)$ for $\zeta(2s)-1$ and $(2)$ for $\zeta(s)-1$, we get
$$
\begin{align}
\frac{\zeta(s)-1}{\zeta(2s)-1}
&\ge\frac{2^{-{\large s}}+\frac3{s-1}3^{-{\large s}}}{\frac{2s+1}{2s-1}2^{-2{\large s}}}\\
&=2^{\large s}\frac{2s-1}{2s+1}+\left(\frac43\right)^{\large s}\frac{6s-3}{(2s+1)(s-1)}\\[3pt]
&=2^{\large s}\left(1-\frac2{2s+1}\right)+\left(\frac43\right)^{\large s}\left(\frac4{2s+1}+\frac1{s-1}\right)\tag{5}
\end{align}
$$
Since $2^{\large s}\left(1-\frac2{2s+1}\right)$ is an increasing function, for $s\ge1$,
$$
2^{\large s}\left(1-\frac2{2s+1}\right)\ge\frac23\tag{6}
$$
Note that $\dfrac{a^x}{x}\ge e\log(a)$. Therefore,
$$
\begin{align}
\left(\frac43\right)^{\large s}\frac4{2s+1}
&=\sqrt3\left(\frac43\right)^{{\large s}+1/2}\frac1{s+\frac12}\\
&\ge\sqrt3\,e\log\left(\frac43\right)\tag{7}
\end{align}
$$
and
$$
\begin{align}
\left(\frac43\right)^{\large s}\frac1{s-1}
&=\frac43\left(\frac43\right)^{{\large s}-1}\frac1{s-1}\\
&\ge\frac43e\log\left(\frac43\right)\tag{8}
\end{align}
$$
Thus, combining $(5)$, $(6)$, $(7)$, and $(8)$ yields
$$
\begin{align}
\frac{\zeta(s)-1}{\zeta(2s)-1}
&\ge\frac23+\left(\frac43+\sqrt3\right)e\log\left(\frac43\right)\\[3pt]
&=3.06379997671918\tag{9}
\end{align}
$$
whereas $\dfrac{3\sqrt3}{2}=2.59807621135332$
A: 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?
