Let $F$ be the CDF for a Beta distribution, $$F(x)=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\int\_{0}^{x}t^{a-1}(1-t)^{b-1}\,dt$$ with $a,b\geq 1$. Is it true that $$\frac{b}{a+b} \leq\int\_0^1\sqrt{F(x)}dx \leq \frac{2b}{a+b}\,?$$ Numerical simulations very strongly suggest that both inequalities hold, but I have no idea how to prove it.



An inequality involving the beta distribution
You understand that it may be a good time to quit when you start forgetting what used to be your favorite tricks.


Let $u=\frac ab>0$. Let $F\_s$ be the CDF corresponding to $a'=u(1+s),b'=1+s$, so $F\_s'(t)=c\_s t^{u-1}[t^u(1-t)]^s$. Notice that $F\_s(0)=0$, $F\_s(1)=1$ and $\int\_0^1 F\_s=\frac{b}{a+b}$ for all $s$. Now, when $s>0$ (we are interested in $s=b-1$), we have $F\_s<F\_0$ slightly to the right of $0$ and $F\_s>F\_0$ slightly to the left of $1$ (the order of tangency is greater). Thus, the graphs of $F\_s$ and $F\_0$ intersect at least once on $(0,1)$.


If they intersect more than once on $(0,1)$, then, by Rolle, the difference of the derivatives $F\_s'-F\_0'$ has at least three zeroes on $(0,1)$ but that difference is $t^{u-1}\{c\_s[t^u(1-t)]^s-c\_0\}$, which, clearly, has at most two roots on $(0,1)$ (the function $t\mapsto t^u(1-t)$ is unimodal). Thus we have $F\_s\le F\_0$ on $[0,x\_s]$, $F\_s\ge F\_0$ on $[x\_s,1]$ and $F\_s(x\_s)=F\_0(x\_s)=Y$ for some $x\_s\in(0,1)$.


Then, since both $F\_s,F\_0$ are increasing and $\Phi(v)=\sqrt v$ is concave, we get
$$
\Phi(F\_s)-\Phi(F\_0)\le \Phi'(Y)(F\_s-F\_0)
$$
on $[0,1]$ (consider $[0,x\_s]$ and $[x\_s,1]$ separately),
so, since $F\_s$ and $F\_0$ have the same integral, we get $\int\_0^1\Phi(F\_s)\le\int\_0^1 \Phi(F\_0)$.


Now it remains to evaluate $\int\_0^1\sqrt{F\_0}=\int\_0^1 x^{u/2}\,dx=\frac{2}{u+2}=\frac{2b}{a+2b}$.


