We know that it can be proved easily that $(s-1)\zeta(s) = s - s(s-1)\int_1^{\infty} \frac{\{x\}}{x^{s+1}}dx$.
I am trying to use this to prove that $\zeta(s) + (s-1)\zeta'(s) > 1 - (2s-1)\int_1^{\infty} \frac{\{t\}}{t^{s+1}}dt$.
It is clear that we have to differentiate the first expression, but how do I get the inequality? I am just missing it somehow. Any pointers? Thanks
If $s>1$ differentiating your equality gives $$\zeta (s)+(s-1)\zeta '(s)=1-(2s-1)\int _1^\infty \frac {\{ t\} dt}{t^{s}}-\underbrace {s(s-1)}_{>0}\underbrace {\frac {d}{ds}\int _1^\infty \frac {\{ t\} dt}{t^{s}}}_{=:X}$$ so we're done if $X<0$. But this is true because $$X=-\underbrace {\int _1^\infty \frac {\{ t\} \log tdt}{t^{s+1}}}_{>0}$$