Inequality $\Gamma(\alpha+x)\Gamma(\beta+y)\leq C(\alpha; \beta)\Gamma(x+y-1)$

Question

I need to prove $$\Gamma(\alpha+x)\Gamma(\beta+y)\leq C(\alpha; \beta)\Gamma(x+y),$$ for $x>\alpha$ and $y>\beta$ with $0<\alpha,\beta \leq \frac{1}{2}$ constants, and $C(\alpha, \beta)$ is a constant depending on $\alpha$ and $\beta$. I guess this holds from numerical computations and I need the especific case where $\alpha=\frac{1}{6}$ and $\beta=\frac{1}{2}$. It probably has something to do with the fact that $$\Gamma(x)\Gamma(y)\leq \Gamma(x+y-1),$$ for $x,y\geq1$, which I know is true for naturals, as $$\Gamma(x)\Gamma(y)=(x-1)!(y-1)!\leq(x+y-2)!=\Gamma(x+y-1),$$ which can be proven with induction, but I also can't prove for the general case where $x,y\in\mathbb{R}$. Any help, suggestion or proof to either of the inequalities is of great help. Thanks in advance!

Answer 1

This is not true for any positive $\alpha,\beta$ without some further hypothesis on $x$ and $y$. Fix positive $y<\alpha$, and let $x \rightarrow \infty$. Then the left-hand side is a constant times $\Gamma(x+\alpha)$ and the right-hand side is a constant times $\Gamma(x+y)$; so the desired inequality would assert that $\Gamma(x+\alpha)/\Gamma(x+y)$ is bounded. But in fact (using Stirling's asymptotic formula for $\Gamma(z)$, for example) this ratio $\Gamma(x+\alpha)/\Gamma(x+y)$ is asymptotic to $x^{\alpha-y} \rightarrow \infty$ as $x \rightarrow \infty$.

Instead the inequality $$ \Gamma(x) \, \Gamma(y) \leq \Gamma(x+y-1) $$ can be proved for all real $x,y \geq 1$ as follows. Equality holds when either $x=1$ or $y=1$; I claim that otherwise $\Gamma(x) \, \Gamma(y) < \Gamma(x+y-1)$. Fix $y>1$. It is known that the Gamma function is logarithmically convex upwards. Therefore $\Gamma(x+y-1)/\Gamma(x)$ is an increasing function of $x$. Because it equals $\Gamma(y)$ for $x=1$, it exceeds $\Gamma(y)$ for all $x>1$, and we are done.

Answer 2

Following inequality is, by the log-convexity, true $$\Gamma(x)\Gamma(y+\epsilon) \leq \Gamma(\epsilon)\Gamma(x+y) \quad , \quad \forall y \geq 0, x \geq \epsilon > 0 \, .$$ Then substitute $$x \rightarrow x+\alpha \\ y \rightarrow y-\alpha \\ \epsilon = \alpha + \beta \\ \Rightarrow \quad \Gamma(x+\alpha)\Gamma(y+\beta) \leq \Gamma(\alpha+\beta)\Gamma(x+y) \quad , \quad \forall y\geq\alpha, x\geq \beta \, .$$