I think this is false for the uniform distribution on $[0,\tfrac14]$ but the calculations are a bit messy, so please double-check this. In general (if I haven't botched something) you want a distribution which has little or no mass on $[\tfrac12,1]$ and reasonably large values of $f(x)$ for some small $x$ where $F(x)$ is close to $1$.

@Erick Wong, you see, I have shown $U'(m)\geq 0$ for $F=m^a, a\geq 0$. When $a=1$, the distribution is uniform. So I guess there are some errors in your calculations.

@user2966729 You are mistaken. I said uniform distribution on $[0,\tfrac14]$ where $f(x) = 4$ on the support of $f$. This is $F(m) = \min(1,4m)$, which you have not tested.

@Erick Wong, oh sorry and thanks, let me check.

@Erick Wong, I think there are errors in your calculations. When $x\in[0,\frac{1}{4}]$, $U'(x)=2 - 8 x + 18 x^2$, which reaches a minimum at $x^*=\frac{2}{9}$, where $U'(x^*)=\frac{10}{9}$. But thanks for inspirations.

@user2966729 Thanks for checking, but did you take into account the fact that $f(x) = 0$ for $x > \tfrac14$ so that $\int_x^1 = \int_x^{1/4}$?.

For $f=0$, the only negative part in $U'(x)$ is removed, see the second equation in my original post.

The problem is that your integral contains a large negative impulse at $1/4$ which isn't accounted for.

I think it would help to go back to the definition of $U(x)$ for this.

By impulse I mean: your calculation is assuming $f$ is differentiable, which isn't true for this case. We can approximate the bump function by something differentiable, but the new function would have to decrease very sharply at $x=1/4$, so that there is a very large negative value of $f_t$ inside the integral.

Thank you so much for your help. Very good insights. I can assume $f$ is everywhere differentiable for less general results on this problem. Is there any other insights for differentiable assumption?

You missed an $x$ in $(2x-1)f(x)$ in the second equality of $U'(x)$. $$U'(x)=(3x-1)F(x)+(2x-1)xf(x)+\int_x^1 F(t)dt.$$. But that doesn't matter. I see you point there.

For this problem it's ok to restrict to everywhere differentiable functions. Any bounds on $U(x)$ should extend to arbitrary distributions by taking limits (so some $>$ signs might turn into $\ge$ signs, etc.)

Main advice is that restricting to distributions with formulas is very restrictive. E.g. the quantity $\int_x^1 F(t) dt$ can be bounded below by $(1-x) F(x)$ and above by $(1-x)$, and there will exist continuous distributions that come arbitrarily close to those bounds. That's the type of thinking that you need here.

I think the bounds you mention is automatically satisfied. If you draw an arbitrary distribution function on $[0,1]$, and take any point on the distribution function, you will see the facts.

So your point about the bounds is ...? I would like to hear about that.

I need to take a ciggarate for 3 minutes. I haven't smoke for hours. Coming right back. Really thankful.