Prove a function containing integrals is positive From plots I find 
$$
U(x)=\int_x^1 t^{b-1}(1-t)^b dt-(1-2x)(1-x)^{b-1}x^b
+2x\int_0^x t^{b-1}(1-t)^{b-1}dt \geq 0
$$ 
for any $x \in [0,\tfrac12], 0<b<1$, (or rather the plots of $\,\, \Gamma(2b+1)*U(x)$, so that the positiveness is more obvious from plots, where$\,\,\Gamma(\cdot)$ is the Gamma function). But mathematically it is too challenging for me.  
It seems that $U(x)$ has a unique stationary point on $x\in[0,\tfrac12]$. But the solution of the stationary point is not closed. It contains the integral $\int_0^x t^{b-1}(1-t)^{b-1}dt$ so that it becomes intractable if we insert it back to $U(x)$.
Some hints on this problem? Thanks.
 A: It seems that we only need the first integral to dominate the negative term. We use the following general inequality (See for example exercise 7.17 in Apostol - Mathematical analysis):

If $f$ and $g$ are both increasing or decreasing on $[a,b]$, then 
  $$
(b-a)\int_a^b f(t)g(t)\,dt\geq \int_a^b f(t)\,dt\int_a^b g(t)\,dt. 
$$

Now, both $f(t)=t^{b-1}$ and $g(t)=(1-t)^b$ decreases on $[x,1]$, and so
$$
\begin{aligned}
(1-x)\int_x^1 t^{b-1}(1-t)^b\,dt&\geq \int_x^1 t^{b-1}\,dt\int_x^1(1-t)^b\,dt\\
&=\frac{1-x^b}{b}\frac{(1-x)^{b+1}}{b+1}.
\end{aligned}
$$
Dividing by $1-x$ gives
$$
\int_x^1 t^{b-1}(1-t)^b\,dt \geq \frac{1-x^b}{b}\frac{(1-x)^{b}}{b+1}.
$$
It remains to show that
$$
(1-2x)(1-x)^{b-1}x^b\leq \frac{1-x^b}{b}\frac{(1-x)^{b}}{b+1},
$$
or, equivalently
$$
\frac{1-2x}{1-x}\frac{x^b}{1-x^b}\leq \frac{1}{b(b+1)}.
$$
This inequality certainly holds for $x\in[0,1/2]$ and $b\in(0,1)$. I suggest that you do the work here.
PS: I accidentally posted while typing this answer. In the non-finished answer that got posted I used a different approach. I apologize for any inconvenience and/or confusion. 
