a proof for the following Gamma function inequality

Question

Could you please provide or point me to a proof of inequality 5.6.8 found at this site? That is,

$\left|\frac{\Gamma(z+a)}{\Gamma(z+b)}\right| \leq \frac{1}{|z|^{b-a}}$

for $z\in \mathbb{C}$, $a,b\in\mathbb{R}$, and $a≥0, b≥a+1, Re(z)>0$.

Answer 1

Thanks to joriki's link to Google Books, here is the passage in question.

R. B. Paris and D. Kaminski, Asymptotics and Mellin-Barnes Integrals, pp. 33-34.

Answer 2

For real $x>0$, the strict log-convexity of $\Gamma$ (see the end of this answer) implies that for $a\ge0$, $b\ge1$ and $b\ge a$, $$ \frac{\log\Gamma(x+1)-\log\Gamma(x)}{1}\le\frac{\log\Gamma(x+b)-\log\Gamma(x+a)}{b-a}\tag{1} $$ Which translates to $$ \frac{\Gamma(x+a)}{\Gamma(x+b)}\le\frac{1}{x^{b-a}}\tag{2} $$ Simpler, but not as general.