I am trying to prove the following inequality concerning the [Beta Function][1]:
$$
\alpha x^\alpha B(\alpha, x\alpha) \geq 1 \quad \forall 0 < \alpha \leq 1, \ x > 0,
$$
where as usual $B(a,b) = \int_0^1 t^{a-1}(1-t)^{b-1}dt$.

In fact, I only need this inequality when $x$ is large enough, but it empirically seems to be true for all $x$.

The main reason why I'm confident that the result is true is that it is very easy to plot, and I've experimentally checked it for reasonable values of $x$ (say between 0 and $10^{10}$). For example, for $x=100$, the plot is:

[![Plot of the function to be proven greater than 1][2]][2]

Varying $x$, it seems that the inequality is rather sharp, namely I was not able to find a point where that product is larger than around $1.5$ (but I do not need any such reverse inequality).

I know very little about Beta functions, therefore I apologize in advance if such a result is already known in the literature. I've tried looking around, but I always ended on inequalities trying to link $B(a,b)$ with $\frac{1}{ab}$, which is quite different from what I am looking for, and also only holds true when both $a$ and $b$ are smaller than 1, which is not my setting.

I have tried the following to prove it, but without success: the inequality is well-known to be an equality when $\alpha = 1$, and the limit for $\alpha \to 0$ should be equal to 1, too. Therefore, it would be enough to prove that there exists at most one $0 < \alpha < 1$ where the derivative of the expression to be bounded vanishes. This derivative can be written explicitly in terms of the [digamma function][3] $\psi$ as:
$$
x^\alpha B(\alpha, x\alpha) \Big(\alpha \psi(\alpha) - (x+1)\alpha\psi((x+1)\alpha) + x\alpha \psi(x\alpha) + 1 + \alpha \log x  \Big).
$$
Dividing by $x^\alpha B(\alpha, x\alpha) \alpha$, this becomes
$$
-f(\alpha) + \frac{1}{\alpha} + \log x,
$$
where $f(\alpha) = -\psi(\alpha) + (x+1)\psi((x+1)\alpha) - x \psi(x\alpha)$ is, as proven [by Alzer and Berg][4], Theorem 4.1, a completely monotonic function. Unfortunately, the difference of two completely monotonic functions (such as $f(\alpha)$ and $\frac{1}{\alpha} + C$) can vanish in arbitrarily many points, therefore this does not allow to conclude.

Many thanks in advance for any hint on how to get such a bound!


  [1]: https://en.wikipedia.org/wiki/Beta_function
  [2]: https://i.sstatic.net/UiRCf.png
  [3]: https://en.wikipedia.org/wiki/Digamma_function
  [4]: http://web.math.ku.dk/~berg/manus/comabnew.pdf