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.
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):
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.