Inequality $\Gamma(\alpha+x)\Gamma(\beta+y)\leq C(\alpha; \beta)\Gamma(x+y-1)$ 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!
 A: 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.
A: 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
$$x \rightarrow x+\alpha \\
y \rightarrow y-\alpha \\
\epsilon = \alpha + \beta \\
\Rightarrow \quad \Gamma(x+\alpha)\Gamma(y+\beta) \leq \Gamma(\alpha+\beta)\Gamma(x+y) \quad , \quad \forall y\geq\alpha, x\geq \beta \, .$$
