lower bound for a function of $\Gamma$ 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!
 A: Note that the cases for $x=0$ and $x=-1/2$ are trivial.
Next, we can write for $0<x<1/2$
$$\begin{align}
\frac{\Gamma(1-2x)\Gamma(1+x)}{\Gamma(1-x)}&=x\frac{\Gamma(1-2x)\Gamma(x)}{\Gamma(1-x)}\\\\
&=xB(1-2x,x)\\\\
&=x\int_0^1 t^{-2x}(1-t)^{x-1}\,dt\\\\
&\ge x\int_0^1(1-t)^{x-1}\\\\
&=1
\end{align}$$
as was to be shown!

For $-1/2<x<0$, we let $y=-x$ and write 
$$\begin{align}
\frac{\Gamma(1-2x)\Gamma(1+x)}{\Gamma(1-x)}&=\frac{\Gamma(1+2y)\Gamma(1-y)}{\Gamma(1+y)}\\\\
&=2yB(2y,1-y)\\\\
&=2y\int_0^1 t^{2y-1}(1-t)^{-y}\,dt\\\\
&\ge 2y\int_0^1 t^{2y-1}\,dt\\\\
&=1
\end{align}$$
as was to be shown!
A: 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:
$$ \log h(x) = \sum_{n\geq 2}\color{red}{\frac{\zeta(n)}{n}(2^n+(-1)^n-1)}\, x^n.\tag{4}$$
