Inequality of logarithm of the tail of the Euler product I would want to know if this inequality holds: Let $x$ be a positive integer and $b>1$ be a real, then
$$
\sum_{p> x}\log(1-p^{-b})\ge-\frac{b}{x},
$$
where the sum is over all prime $p\ge x$.
EDIT: Actually, I would want to see if this inequality is true when: Fix $b>1$ be a real. For a large positive integer $x$, that inequality holds. 
 A: We have $$\sum_{p>x}\log\left(1-\frac{1}{p^{b}}\right)=\sum_{n>x}\left(\pi\left(n\right)-\pi\left(n-1\right)\right)\log\left(1-\frac{1}{n^{b}}\right)=\sum_{n>x}\pi\left(n\right)\left(\log\left(1-\frac{1}{n^{b}}\right)-\log\left(1-\frac{1}{\left(n+1\right)^{b}}\right)\right)=
 $$ $$=-b\sum_{n>x}\pi\left(n\right)\int_{n}^{n+1}\frac{1}{t\left(t^{b}-1\right)}dt=-b\int_{x}^{\infty}\frac{\pi\left(t\right)}{t\left(t^{b}-1\right)}dt
 $$ and we can observe that, if $b\geq2
 $ $$\int_{x}^{\infty}\frac{\pi\left(t\right)}{t^{b}-1}dt=O\left(\int_{x}^{\infty}\frac{1}{\log\left(t\right)t^{b-1}}dt\right)=O\left(\frac{x^{2-b}}{\log\left(x\right)}\right)=o\left(1\right)
 $$ so $$-b\int_{x}^{\infty}\frac{\pi\left(t\right)}{t\left(t^{b}-1\right)}dt\geq-\frac{b}{x}\int_{x}^{\infty}\frac{\pi\left(t\right)}{t^{b}-1}dt\geq-\frac{b}{x}.
 $$ If $1<b<2
 $ the inequality doesn't hold, because $$\int_{x}^{\infty}\frac{\pi\left(t\right)}{t\left(t^{b}-1\right)}dt\approx\int_{x}^{\infty}\frac{1}{t^{b}\log\left(t\right)}dt\approx\frac{1}{x^{b-1}}
 $$ and so we have a denominator smaller than $x
 .$
