# a proof for the following Gamma function inequality

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 $a≥0, b≥a+1, Re(z)>0$.

• At that link it has additional assumptions which you left out: $a \ge 0$, $b \ge a + 1$, $\text{Re}(z) > 0$. – Robert Israel May 4 '12 at 4:03
• The (i) link to the right of the inequality gives a reference to Paris and Kaminski (2001): dlmf.nist.gov/bib/P#bib1827 – Robert Israel May 4 '12 at 4:08
• Also you got the left side wrong: it should be $\left| \dfrac{\Gamma(z+a)}{\Gamma(z+b)}\right|$ – Robert Israel May 4 '12 at 4:31
• Ah thanks for that! – yep May 4 '12 at 5:11
• The book is available at Google Books, but p. 34, to which the citation points, isn't shown (for me). – joriki May 4 '12 at 5:37

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.

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.

• Thanks - I think this is useful to have as well. – yep May 6 '12 at 3:04