Let $\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$. We have $\frac{\zeta'}{\zeta}(s) = \sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}$ for $s>1$, where $\Lambda$ stands for the von Mangoldt function (http://en.wikipedia.org/wiki/Von_Mangoldt_function). It is easy to prove that there exists a constant $A$ such that $-\frac{\zeta'}{\zeta}(s) \leq \frac{1}{s-1} + A$ for every $s>1$. By plotting $-\frac{\zeta'}{\zeta}(s)-\frac{1}{s-1}$ or $-\frac{\zeta'}{\zeta}(s)\cdot(s-1)$ on any computation program, it seems fair to conjecture that we can choose $A=0$. Unfortunately, I am unable to prove this. Is it a well-known inequality? Does anyone know how to prove this, if true?
I tried to use the formula $-\frac{\zeta'}{\zeta}(s) = \frac{1}{s-1} + 1 + s \int_1^\infty \frac{\psi(t)-t}{t^{s+1}} \mathrm{d}t$, where $\psi(x) = \sum_{n \leq x} \Lambda(n)$, but the erractic behavior of $\psi$ makes it useless (and any Chebyshev-type inequality does give $A>0$).
Thanks in advance.
Ok, I succeeded in proving this. From the equality $\frac{\zeta'}{\zeta}(s)+\frac{1}{s-1} =\frac{1-(2s-1)I(s)-s(s-1)I'(s)}{\textrm{something positive}}$ (see my comment below my question), since $I'(s)<0$, we only have to prove thaht $1-(2s-1)I(s)>0$. Since $I(s)$ can be expressed in terms of $\zeta$, it is equivalent to an inequality only involving $\zeta$, that is, $\zeta(s) > \frac{s^2}{(s-1)(2s-1)}$. This is easy to prove, because we have refined lower bounds for $\zeta$ thanks to the comparaison between series and integrals.