Gfrktyl

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.

You can get away with the usual distribution function mumbo-jumbo. The general lemma is as follows:

Let $mu,nu$ be non-negative measures and $f,g$ be non-negative functions such that there exists $s_0>0$ with the property that $mu{f>s}ge nu{g>s}$ for $sle s_0$ and the reverse inequality holds for $sge s_0$. Suppose also that $int f^q,dmu=int g^q,dnu<+infty$ for some $q>0$. Then, as long as the integrals in question are finite, we have $int f^p,dmuge int g^p,dnu$ for $0<ple q$ and the reverse inequality holds for $pge q$.

The proof of the lemma is rather straightforward. Let $ple q$ (that is the case you are really interested in)
$$
int f^p,dmu-int g^p,dnu=pint_0^infty s^p[mu{f>s}-nu{g>s}]frac{ds}s
\
=pint_0^infty [s^p-s_0^{p-q}s^q][mu{f>s}-nu{g>s}]frac{ds}sge 0,.
$$

Now we use it with $f(t)=t(1-t)^x$, $dmu=frac{dt}{t(1-t)}$ on $(0,1)$, $g(t)=t$, $dnu=frac{dt}{t}$ on $(0,frac1x)$. Since the maximum of $t(1-t)^x$ is attained at $t=frac{1}{x+1}$, we see that the function $smapsto mu{f>s}$ drops to $0$ before the function $smapsto nu{g>s}$. Also, the first function has larger in absolute value negative derivative than the second one for each value of $s$ where it is still positive. To see it, notice that the set where $f>s$ is an interval $(u,v)=(u(s),v(s))$ that shrinks as $s$ increases and the left end $u$ of this interval satisfies
$$
duleft(frac 1u-frac x{1-u}right)=frac{ds}s,,
$$
so trivially
$$
frac{du}{u(1-u)}ge frac{du}u>frac {ds}s
$$
The right end moving to the left can only increase the decay speed. Finally, for $q=1$, the integrals are equal (which also shows that the graphs of the distribution functions must indeed intersect), so for $0<ple 1$ (which plays the role of $alpha$), we have the desired inequality.

One can also use Jensen's inequality. Let (for $sigma>0$) $G_sigma$ denote a random variable with $Gamma(1,sigma)$-distribution, i.e. having Lebesgue density
$$f_sigma(t)=frac{t^{sigma-1}}{Gamma(sigma)} e^{-t};1_{(0,infty)}(t);,$$
then $mathbb{E}(G_sigma)=sigma$.
Since $alphain (0,1)$ the functions $tmapsto t^alpha$ resp. $tmapsto t^{1-alpha}$ on $mathbb{R}_+$ are concave. By Jensen's inequality
$$frac{Gamma(alpha+alpha x)}{Gamma(alpha x)}=mathbb{E}(G_{xalpha}^alpha)leq left(mathbb{E}(G_{xalpha})right)^alpha=(xalpha)^{alpha}$$

and
$$frac{1}{Gamma(alpha)}=mathbb{E} G_alpha^{1-alpha}leqleft(mathbb{E}(G_{alpha})right)^{1-alpha}=frac{1}{alpha^{alpha-1}}$$
Using that gives
$$B(alpha,x alpha)=frac{Gamma(alpha),Gamma(xalpha)}{Gamma(alpha +xalpha)}geq frac{Gamma(alpha)}{alpha^alpha x^alpha}geq frac{Gamma(alpha)}{alpha,Gamma(alpha),x^alpha}=frac{1}{alpha x^alpha},$$
as desired.

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.