Let $F:[0,1]\to [0,1]$ be an increasing function with $F(0)=0, F(1)=1$. Define $A(x)=1-\int_{x}^1 F(t) dt$. I am trying to approximately minimize the following ratio across all $F$ (i.e find a lower bound on the ratio).
$$\frac{A(A(0))}{1-\int_{0}^1 F(t)^2 dt} = \frac{1-\int_{1-\int_{0}^1 F(r)dr }^1 F(t) dt }{1-\int_{0}^1 F(t)^2 dt}$$
My attempt:
I tried approximating $F$ using Stone–Weierstrass theorem: Letting $f_i = F(\frac{i}{5})$, then
$$F(x) \approx \sum_{i=0}^5 f_i {5 \choose i}x^i (1-x)^{5-i}$$
Then I evaluated the ratio (on Mathematica), and threw that into an optimizer subject to $f_0=0, f_5=1$ and $f_i \leq f_{i+1}$ constraints. The minimum seems to be $\approx 0.930147$. Replacing $5$ with $8$ above gets $\approx 0.917706$
However, I don't see how to generalize this for higher powers analytically using Bernstein polynomials.
I will use a function $F:(0,1) \to (0,1)$ where $\lim_{t \to 0^+} F(t)=0$ and $\lim_{t \to 1^-} F(t)=1.$
Let $F_r(t)=e^{\frac{1/r}{\log t}}$ where you'd like to minimize the following ratio over $r$:
$$E(r)= \frac{1-\int_{1-\frac{2K_1\big(\sqrt{\frac{1}{r}}\big)}{\sqrt{r}}}^1e^{\frac{1/r}{\log t}}dt}{1-\frac{\sqrt{2}K_1\big( \sqrt{\frac{2}{r}} \big)}{\sqrt{r}}} $$
We're looking for $\text{min}(E(r))\approx E(85) \approx 0.911078522836.$ (this minimum was found numerically using Mathematica).
Here $K_1$ is a modified Bessel function of the second kind.
This may not be the minimum value across any choice of $F$ but it improves the minimum in your post and I hope it helps.
Note that I've suppressed some calculations in the expression $E(r)$ to make it more concise. I can add in all of the steps if needed.