Proving that $B(2\alpha,1-\alpha)\le\frac1{\alpha(1-\alpha)}$

Question

Let $0<\alpha<\frac12$, I cannot prove that

$$ B(2\alpha,1-\alpha)\le\frac1{\alpha(1-\alpha)} $$

where $$ B(p,q):=\int_0^1t^{p-1}(1-t)^{q-1}\,dt=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)} $$ is the Euler Beta function, defined as above for $\Re p,\Re q>0$; the last relation, with the Gamma function holds for these value of $p,q$, as is well known (but don't care about complex numbers, here $\alpha\in\Bbb R, 0<\alpha<1/2$).

I tried in many ways, e.g. like this \begin{align*} B(2\alpha,1-\alpha) &=\frac{\Gamma(2\alpha)\Gamma(1-\alpha)}{\Gamma(\alpha+1)}\\ &=\frac{\Gamma(2\alpha)\Gamma(1-\alpha)}{\alpha\Gamma(\alpha)} \end{align*} then I used the duplication formula $$ \Gamma(2\alpha)=\frac{\Gamma(\alpha)\Gamma(\alpha+\frac12)}{2^{2\alpha-1}\sqrt{\pi}} $$ but doesn't seem leading nowhere useful.

I tried even with a geometrical approach, looking at the real plot of $\Gamma$ and remeberig that $\gamma(1/2)=\sqrt{\pi},\Gamma(1)=\Gamma(2)=1$, but neither this was helpful.

Answer 1

For any $\alpha\in\left(0,\frac{1}{2}\right)$ we have

$$ B(2\alpha,1-\alpha)=\int_{0}^{1}(1-t)^{2\alpha-1}t^{-\alpha}\,dt =\frac{\Gamma(2\alpha)\Gamma(1-\alpha)}{\alpha\Gamma(\alpha)}$$ hence it is enough to show that $\Gamma(2\alpha)\,\Gamma(2-\alpha)\leq\Gamma(\alpha)$ holds over $\left(0,\frac{1}{2}\right)$.
We may notice that

$$ \frac{d}{d\alpha}\log\left(\frac{\Gamma(2\alpha)\Gamma(2-\alpha)}{\Gamma(\alpha)}\right) = \sum_{n\geq 0}\left(\frac{1}{n+2-\alpha}+\frac{1}{n+\alpha}-\frac{2}{n+2\alpha}\right) $$ hence by differentiating again we get that $\frac{\Gamma(2\alpha)\Gamma(2-\alpha)}{\Gamma(\alpha)}$ is a log-convex function on $\left(0,\frac{1}{2}\right)$.
A positive log-convex function is also convex, hence the supremum of $\frac{\Gamma(2\alpha)\Gamma(2-\alpha)}{\Gamma(\alpha)}$ over $\left(0,\frac{1}{2}\right)$ is attained at the boundary and the claim follows from $$ \lim_{\alpha\to 0^+}\frac{\Gamma(2\alpha)\Gamma(2-\alpha)}{\Gamma(\alpha)}=\lim_{\alpha\to \frac{1}{2}^-}\frac{\Gamma(2\alpha)\Gamma(2-\alpha)}{\Gamma(\alpha)}=\frac{1}{2}.$$

$$ \forall \alpha\in\left(0,\frac{1}{2}\right),\qquad B(2\alpha,1-\alpha)<\frac{1}{\color{red}{2}\alpha(1-\alpha)}.$$

Additionally, $B(2\alpha,1-\alpha)\approx \frac{1}{2\alpha(1-\alpha)}+\alpha-\frac{1}{2}$ is a quite accurate approximation over $\left(0,\frac{1}{2}\right)$.