TITLE: lower bound for a function of $\Gamma$
QUESTION [7 upvotes]: I would like to show that $$\frac{\Gamma(1-2x)\Gamma(1+x)}{\Gamma(1-x)}\geq 1$$ for real $x$ such that $|x|\leq \frac12$, where $\Gamma$ is the usual gamma function.
I looked at the derivatives and went a long road to prove this. I though somehow believe there should be a simpler way. I appreciate any hints or comments. Many thanks!

REPLY [2 votes]: Since
$$ \Gamma(1+t)=e^{-\gamma t}\prod_{n\geq 1}\left(1+\frac{t}{n}\right)^{-1}e^{t/n} \tag{1}$$
we have:
$$ h(x)=\frac{\Gamma(1-2x)\,\Gamma(1+x)}{\Gamma(1-x)}=\prod_{n\geq 1}\frac{\left(1-\frac{x}{n}\right)}{\left(1-\frac{2x}{n}\right)\left(1+\frac{x}{n}\right)}\tag{2}$$
and it is straightforward to check that $h(x)$ is a log-convex function on $I=\left(-\frac{1}{2},\frac{1}{2}\right)$.
Since $h'(0)=0$, $h(0)=1$ is a local minimum for $h(x)$ on $I$.

We may even compute the Taylor series of $\log h(x)$. Since:
$$ \sum_{n\geq 2}\frac{\zeta(n)}{n} t^n = -\gamma t+\log\Gamma(1-t)\tag{3} $$
we have:
$$ \log h(x) = \sum_{n\geq 2}\color{red}{\frac{\zeta(n)}{n}(2^n+(-1)^n-1)}\, x^n.\tag{4}$$