Cfrtjryk

I am trying to prove the following inequality concerning the Beta Function:
$$
alpha x^alpha B(alpha, xalpha) geq 1 quad forall 0 < alpha leq 1, x > 0,
$$
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:

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 $psi$ as:
$$
x^alpha B(alpha, xalpha) Big(alpha psi(alpha) - (x+1)alphapsi((x+1)alpha) + xalpha psi(xalpha) + 1 + alpha log x Big).
$$
Dividing by $x^alpha B(alpha, xalpha) alpha$, this becomes
$$
-f(alpha) + frac{1}{alpha} + log x,
$$
where $f(alpha) = -psi(alpha) + (x+1)psi((x+1)alpha) - x psi(xalpha)$ is, as proven by Alzer and Berg, 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!

[EDIT]: As pointed out in the comments, the link to the paper of Alzer and Berg pointed to the wrong version, I have corrected the link.

This is an attempt to strengthen your claim.

If $x$ is large then $B(x,y)sim Gamma(y)x^{-y}$ and hence
$$B(alpha x,alpha)sim Gamma(alpha)(alpha x)^{-alpha};$$
where $Gamma(z)$ is the Euler Gamma function.

On the other hand, for small $alpha$, we have the expansion
$$Gamma(1+alpha)=1+alphaGamma'(1)+mathcal{O}(alpha^2).$$
Since $alphaGamma(alpha)=Gamma(1+alpha)$, it follows that
$$Gamma(alpha)sim frac1{alpha}-gamma+mathcal{O}(alpha)$$
where $gamma$ is the Euler constant.

We may now combine the above two estimates to obtain
$$alpha x^{alpha}B(alpha x,alpha)sim alpha x^{alpha}left(frac1{alpha}-gammaright)(alpha x)^{-alpha}=left(frac1{alpha}-gammaright)alpha^{1-alpha}geq1$$
provided $alpha$ is small enough. For example, $0<alpha<frac12$ works.