Proving an inequality, involving the Riemann-$\zeta$-function I'm studying a book, which wants to prove that for some constants $C,\kappa$, we have
$$\left|\frac{\zeta ''(s)\zeta (s) -2\zeta '(s)}{\zeta (s)^2}\right|\le C*|t|^\kappa~~~~,\text{for }s=\sigma+it,~|t|\ge 1, \sigma >1.$$
It has proved so far, that for every $m\in \mathbb{N}_0,$ we have constants $C_m$ with
$$|\zeta^{(m)}(s)|\le C_m|t|~~~\text{, for }|t|\ge 1,\sigma>1$$
and for some $\delta>0$ we have:
$$|\zeta(s)|\ge\delta|t|^{-4}~~~\text{, for }|t|\ge 1,\sigma>1.$$
It's obvious that the first inequality holds for e.g. $\kappa=9$. But my book says that the first inequality is true for $\kappa=5,$ too and I fail to see how we can prove this using these inequalities.
How can I prove the first inequality for $\kappa =5$? (is it even possible?)
 A: I think we can get a better approximation of $1/\zeta^{2}\left(s\right)
 $. Take $\epsilon>0
 $ and $\sigma\geq1+\epsilon>1
 $. We recall that if $\sigma>1
 $ we have $$\frac{1}{\zeta\left(s\right)}=\sum_{n\geq1}\frac{\mu\left(s\right)}{n^{s}} \tag{1}
 $$ where $\mu\left(s\right)
 $ is the Mobius function. Now if we consider the partial sum of $(1)
 $ we have, by partial summation, $$ \sum_{n\leq N}\frac{\mu\left(n\right)}{n^{s}}=M\left(N\right)N^{-s}+s\int_{1}^{N}M\left(u\right)u^{-s-1}du
 $$ where $M\left(N\right)=\sum_{n\leq N}\mu\left(n\right)
 $ is the Mertens function. By PNT, we know that $M\left(N\right)=o\left(N\right)
 $ as $N\rightarrow\infty
 $, then $$\left|\frac{1}{\zeta\left(s\right)}\right|=O\left(\left|s\right|\int_{1}^{\infty}u^{-\sigma}du\right)=O\left(\frac{\left|s\right|}{1-\sigma}\right)
 $$ then we have that exists some $C>0
 $ such that $$\left|\frac{1}{\zeta\left(s\right)}\right|\leq C\left|t\right|
 $$ and so $$\left|\frac{\zeta''\left(s\right)\zeta\left(s\right)-2\zeta'\left(s\right)}{\zeta^{2}\left(s\right)}\right|\leq C_{1}\left|t\right|^{3}.
 $$
