I am trying to show that $$\int_0^1F_n(x) dx \leq \int_0^1F_{n+1}(x) dx$$ when $$F_n(x) = (1-(1-F_{n-1}(x))^c)^c$$ and $F_0(x) = x$ and $n$ and $c$ are integers, $n\geq 1$ and $c \geq 2$

Note that $F$ performs a "minimax type opertation". Starting with the uniform distribution, it produced the cdf of the distribution of the maximum of $c$ draws from the minimum of $c$ draws from the distribution before it.

Here is a list of things I know that might be helpful:

1) $$\int_0^1 F_0(x) dx = \int_0^1 x dx < \int (1-(1-x)^c)^c dx= \int_0^1 F_1(x) dx$$

2)

The distribution $F_n$ converges to Dirac mass at the unique solution to $ L= (1-(1-L)^c)^c$, and $\frac{1}{2}^c< L <\frac{1}{2}$ The integral of $F_n$ converges to $1-L$

3)

The integrals of the sequence defined by $F$ are identical to those defined by related functions. $$\int_0^1 F_n dx = \int_0^1 G_n dx = \int_0^11-F_n^{-1} dx= \int_0^1 1-G_n^{-1}dx = $$ $$\int_0^1 1-(F_n(x^c))^{\frac{1}{c}} dx= \int_0^1 (1-F_n(x^c))^c dx$$ where $G_n = (1-(1-G_{n-1}^{\frac{1}{c}})^{\frac{1}{c}} $

4)

$$\int_0^1 F_{n+1}(x)\,dx = \int_0^1F_n(x)dF^{-1}_1(x)dx$$ (and various other ways of rewriting $\int F_i$ in terms of $\int F_j$)

5)

The sequence operator $F= (1-(1-x)^c)^c$ is really two iterations of the simpler sequence operator $(1-x)^c$

6)

The derivative of $F_n$ wrt $x$ = $$f_n = \displaystyle\prod_{k=1}^{n-1}c^2\left(1-\left(1-F_k\right)^c\right)^{c-1}(1-F_k)^{c-1}$$ $$=c^2\left(1-\left(1-F_{n-1}\right)^c\right)^{c-1}(1-F_{n-1})^{c-1}f_{n-1} $$

Does anyone have any ideas?