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?

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 $\dfrac{3\sqrt3}{2}=2.59807621135332$

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?