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Use Jensen's inequality and induction on $n$.

Let $\phi_c(z) = (1- (1 - z)^c)^c$ , so that $\phi_c(F_i) = F_{i+1}$. As you noted, $\phi = f \circ f$, where $f(z) = (1-z)^c$. $f$ is convex for $0 \leq z \leq 1$, since $\frac{d^2f}{dz^2} = c^2(1-z)^{c-2} \geq 0$. By chain rule, $\phi$ is convex when $0 \leq z \leq 1 - \frac{1}{(c+1)^{(1/c)}}$, so the rest of this argument doesn't work.

Now apply Jensen's inequality to get $\phi(\int F_i) \leq \int F_{i+1}$ provided $0 \leq F_i(x) \leq 1$, which holds for all $0 \leq x \leq 1$. For $z > \frac{3 - \sqrt(5)}{2}$, $\phi_c(z) > z$ for every $c \geq 2$. So since $\int F_0 = \frac{1}{2}$, we have $\int F_0 \leq \phi(\int F_0)$, and the result follows by induction.

Use Jensen's inequality and induction on $n$.

Let $\phi_c(z) = (1- (1 - z)^c)^c$ , so that $\phi_c(F_i) = F_{i+1}$. As you noted, $\phi = f \circ f$, where $f(z) = (1-z)^c$. $f$ is convex for $0 \leq z \leq 1$, since $\frac{d^2f}{dz^2} = c^2(1-z)^{c-2} \geq 0$. By chain rule, $\phi$ is convex when $0 \leq z \leq 1 - \frac{1}{(c+1)^{(1/c)}}$, so the rest of this argument doesn't work.

Now apply Jensen's inequality to get $\phi(\int F_i) \leq \int F_{i+1}$ provided $0 \leq F_i(x) \leq 1$, which holds for all $0 \leq x \leq 1$. For $z > \frac{3 - \sqrt(5)}{2}$, $\phi_c(z) > z$ for every $c \geq 2$. So since $\int F_0 = \frac{1}{2}$, we have $\int F_0 \leq \phi(\int F_0)$, and the result follows by induction.

Jensen's inequality could still be used on $f$, but that doesn't seem to get any closer to a solution.

Use Jensen's inequality and induction on $n$.

Let $\phi_c(z) = (1- (1 - z)^c)^c$ , so that $\phi_c(F_i) = F_{i+1}$. As you noted, $\phi = f \circ f$, where $f(z) = (1-z)^c$. $f$ is convex for $0 \leq z \leq 1$, since $\frac{d^2f}{dz^2} = c^2(1-z)^{c-2} \geq 0$. By chain rule, $\phi$ is convex when $0 \leq z \leq 1 - \frac{1}{(c+1)^c}$, so the rest of this argument doesn't work.

Now apply Jensen's inequality to get $\phi(\int F_i) \leq \int F_{i+1}$ provided $0 \leq F_i(x) \leq 1$, which holds for all $0 \leq x \leq 1$. For $z > \frac{3 - \sqrt(5)}{2}$, $\phi_c(z) > z$ for every $c \geq 2$. So since $\int F_0 = \frac{1}{2}$, we have $\int F_0 \leq \phi(\int F_0)$, and the result follows by induction.

Use Jensen's inequality and induction on $n$.

Let $\phi_c(z) = (1- (1 - z)^c)^c$ , so that $\phi_c(F_i) = F_{i+1}$. As you noted, $\phi = f \circ f$, where $f(z) = (1-z)^c$. $f$ is convex for $0 \leq z \leq 1$, since $\frac{d^2f}{dz^2} = c^2(1-z)^{c-2} \geq 0$. By chain rule, $\phi$ is convex when $0 \leq z \leq 1 - \frac{1}{(c+1)^{(1/c)}}$, so the rest of this argument doesn't work.

Now apply Jensen's inequality to get $\phi(\int F_i) \leq \int F_{i+1}$ provided $0 \leq F_i(x) \leq 1$, which holds for all $0 \leq x \leq 1$. For $z > \frac{3 - \sqrt(5)}{2}$, $\phi_c(z) > z$ for every $c \geq 2$. So since $\int F_0 = \frac{1}{2}$, we have $\int F_0 \leq \phi(\int F_0)$, and the result follows by induction.

Use Jensen's inequality and induction on $n$.

Let $\phi_c(z) = (1- (1 - z)^c)^c$ , so that $\phi_c(F_i) = F_{i+1}$. As you noted, $\phi = f \circ f$, where $f(z) = (1-z)^c$. $f$ is convex for $0 \leq z \leq 1$, since $\frac{d^2f}{dz^2} = c^2(1-z)^{c-2} \geq 0$. By chain rule, $\phi$ is convex when $0 \leq z \leq 1$.

Now apply Jensen's inequality to get $\phi(\int F_i) \leq \int F_{i+1}$ provided $0 \leq F_i(x) \leq 1$, which holds for all $0 \leq x \leq 1$. For $z > \frac{3 - \sqrt(5)}{2}$, $\phi_c(z) > z$ for every $c \geq 2$. So since $\int F_0 = \frac{1}{2}$, we have $\int F_0 \leq \phi(\int F_0)$, and the result follows by induction.

Use Jensen's inequality and induction on $n$.

Let $\phi_c(z) = (1- (1 - z)^c)^c$ , so that $\phi_c(F_i) = F_{i+1}$. As you noted, $\phi = f \circ f$, where $f(z) = (1-z)^c$. $f$ is convex for $0 \leq z \leq 1$, since $\frac{d^2f}{dz^2} = c^2(1-z)^{c-2} \geq 0$. By chain rule, $\phi$ is convex when $0 \leq z \leq 1 - \frac{1}{(c+1)^c}$, so the rest of this argument doesn't work.

Now apply Jensen's inequality to get $\phi(\int F_i) \leq \int F_{i+1}$ provided $0 \leq F_i(x) \leq 1$, which holds for all $0 \leq x \leq 1$. For $z > \frac{3 - \sqrt(5)}{2}$, $\phi_c(z) > z$ for every $c \geq 2$. So since $\int F_0 = \frac{1}{2}$, we have $\int F_0 \leq \phi(\int F_0)$, and the result follows by induction.