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 $\int_0^1\sqrt{F_0}=\int_0^1 x^{u/2}\,dx=\frac{2}{u+2}=\frac{2b}{a+2b}$.

The first inequality was proved in the comment by user fedja.

Here are some observations concerning the second inequality.

1. Using the Cauchy--Schwarz inequality and fedja's observation that $\int_0^1 F=\frac b{a+b}$, we get $$\int_0^1\sqrt F\le\sqrt{\int_0^1 F}=\sqrt{\frac b{a+b}}\le\frac{2b}{a+b}$$ if $$r:=\frac a{a+b}\le\frac34.$$ So, the hard case for the second inequality is when $r\in(3/4,1)$.

2. For $b=1$, a simple calculation shows that the two sides of the second inequality are asymptotically equivalent to each other. So, the constant factor $2$ on the right-hand side of the second inequality is optimal.

3. If $a,b\to\infty$, then, by the compactness of the interval $[0,1]$, without loss of generality $r$ converges to some $r_*\in[0,1]$. Then (using e.g. a law of large numbers) it is easy to see that $F(x)\to1(x>r_*)$ for each $x\in[0,1]\setminus\{r_*\}$. It follows that the left- and right-hand sides of the second inequality converge to $1-r_*$ and $2(1-r_*)$, respectively. So, the second inequality "holds in the limit" if $\limsup r<1$.

4. So, the hardest case appears to be when $r_*=1$, when the "phase transition" between $1$ and $1/2$ occurs for the ratio of left- and right-hand sides of the second inequality, as illustrated in the following graph of this ratio (click on it to enlarge):

Once this case is handled, the rest can in principle be done by the interval method.