Let $c>1$ be any fixed real exponent and let us denote, for any $x\in I:=[0,1]$, $f(x):=(1-x)^c$, and $g(x):=1-x^{\frac{1}{c}}$. So $f$ is a strictly decreasing homeomorphism of $I$ into itself, with inverse map $g$ and with the interior fixed point $0 < L< 1/2$. Also, since all even-order iterated of $f$ are strictly increasing, for any $x\in I$ and for any $n\in\mathbb{N}$, we have $$x\le L\quad\mathrm{iff}\quad f(x)\ge x\quad\mathrm{iff}\quad F_1(x)\le x \quad\mathrm{iff}\quad F_{n+1}(x)\le F_n(x)\\ .$$ The sets $$A:=\{(x,y)\in I^2\\ : \\ F_{n+1}(x)\le y \le F_n(x)\}$$ and $$B:=\{(x,y)\in I^2\\ : \\ F_{n+1}(x)\ge y \ge F_n(x)\}$$ are therefore included in the squares $[0,L]^2$, respectively $[L,1]^2$, so we have $$-\int_0^L(F_{n+1}-F_n)dx=|A|$$ and $$\int_L^1(F_{n+1}-F_n)dx=|B|\\ .$$ Moreover, $(x,y)\in A$, that is $ F_{n+1}(x)\le y \le F_n(x)$, if and only if $$F _ {n+1}(f(x))=f(F _ {n+1}(x))\ge f(y) \ge f(F_ n(x))=F_ n(f(x))\\ ,$$ that is, $(f\times f)(x,y):=(f(x), f(y))\in B$. So $A=(g\times g)(B)\\ .$ The map $g\times g$ has Jacobian determinant $g'(x)g'(y)=\frac{1}{c^2}(xy)^{\frac{1}{c}-1}\\ ,$ and since $x\ge L$ and $y\ge L$ in $B$ $$|A|=|(g\times g)(B)|=\int_{B} g'(x)g'(y)dxdy\le \left( \frac{L^{\frac{1}{c}-1}}{c}\right)^2|B|\\ . $$

Let's prove that the factor in front of $|B|$ is less than or equal to $1$. So we wish to show that $$L^{\frac{1}{c}-1}\le c\\ . $$ Recalling that $L=(1-L)^c$, that is, $c=\frac{\log L}{\log (1-L)}$, this reduces to

$$\frac{(1-L)\log(1- L)} {L\log L}\le 1 $$

which is indeed true exactly for $L\le 1/2$. So we have $$\int_0^1 F_{n+1}dx -\int_0^1 F_n dx = \int_0^L(F_{n+1}-F_n)dx + \int_L^1(F_{n+1}-F_n)dx=$$ $$= -|A|+|B|\ge 0\\, $$ as required.

Let $c>1$ be any fixed real exponent and let us denote, for any $x\in I:=[0,1]$, $f(x):=(1-x)^c$, and $g(x):=1-x^{\frac{1}{c}}$. So $f$ is a strictly decreasing homeomorphism of $I$ into itself, with inverse map $g$ and with the interior fixed point $0 < L< 1/2$. Also, since all even-order iterated of $f$ are strictly increasing, for any $x\in I$ and for any $n\in\mathbb{N}$, we have $$x\le L\quad\mathrm{iff}\quad f(x)\ge x\quad\mathrm{iff}\quad F_1(x)\le x \quad\mathrm{iff}\quad F_{n+1}(x)\le F_n(x)\, .$$ The sets $$A:=\{(x,y)\in I^2\, : \, F_{n+1}(x)\le y \le F_n(x)\}$$ and $$B:=\{(x,y)\in I^2\, : \, F_{n+1}(x)\ge y \ge F_n(x)\}$$ are therefore included in the squares $[0,L]^2$, respectively $[L,1]^2$, so we have $$-\int_0^L(F_{n+1}-F_n)dx=|A|$$ and $$\int_L^1(F_{n+1}-F_n)dx=|B|\, .$$ Moreover, $(x,y)\in A$, that is $ F_{n+1}(x)\le y \le F_n(x)$, if and only if $$F _ {n+1}(f(x))=f(F _ {n+1}(x))\ge f(y) \ge f(F_ n(x))=F_ n(f(x))\, ,$$ that is, $(f\times f)(x,y):=(f(x), f(y))\in B$. So $A=(g\times g)(B)\, .$ The map $g\times g$ has Jacobian determinant $g'(x)g'(y)=\frac{1}{c^2}(xy)^{\frac{1}{c}-1}\, ,$ and since $x\ge L$ and $y\ge L$ in $B$ $$|A|=|(g\times g)(B)|=\int_{B} g'(x)g'(y)dxdy\le \left( \frac{L^{\frac{1}{c}-1}}{c}\right)^2|B|\, . $$

Let's prove that the factor in front of $|B|$ is less than or equal to $1$. So we wish to show that $$L^{\frac{1}{c}-1}\le c\, . $$ Recalling that $L=(1-L)^c$, that is, $c=\frac{\log L}{\log (1-L)}$, this reduces to

$$\frac{(1-L)\log(1- L)} {L\log L}\le 1 $$

which is indeed true exactly for $L\le 1/2$. So we have $$\int_0^1 F_{n+1}dx -\int_0^1 F_n dx = \int_0^L(F_{n+1}-F_n)dx + \int_L^1(F_{n+1}-F_n)dx=$$ $$= -|A|+|B|\ge 0\,, $$ as required.

Let $c>1$ be any fixed real exponent and let us denote, for any $x\in I:=[0,1]$, $f(x):=(1-x)^c$, and $g(x):=1-x^{\frac{1}{c}}$. So $f$ is a strictly decreasing homeomorphism of $I$ into itself, with inverse map $g$ and with the interior fixed point $0 < L< 1/2$. Also, since all even-order iterated of $f$ are strictly increasing, for any $x\in I$ and for any $n\in\mathbb{N}$, we have $$x\le L\quad\mathrm{iff}\quad f(x)\ge x\quad\mathrm{iff}\quad F_1(x)\le x \quad\mathrm{iff}\quad F_{n+1}(x)\le F_n(x)\\ .$$ The sets $$A:=\{(x,y)\in I^2\\ : \\ F_{n+1}(x)\le y \le F_n(x)\}$$ and $$B:=\{(x,y)\in I^2\\ : \\ F_{n+1}(x)\ge y \ge F_n(x)\}$$ are therefore included in the squares $[0,L]^2$, respectively $[L,1]^2$, so we have $$-\int_0^L(F_{n+1}-F_n)dx=|A|$$ and $$\int_L^1(F_{n+1}-F_n)dx=|B|\\ .$$ Moreover, $(x,y)\in A$, that is $ F_{n+1}(x)\le y \le F_n(x)$, if and only if $$F_n(f(x))=f(F_{n+1}(x))\ge f(y) \ge f(F_n(x))=F_n(f(x))\\ ,$$ that is, $(f\times f)(x,y):=(f(x), f(y))\in B$. So $A=(g\times g)(B)\\ .$ The map $g\times g$ has Jacobian determinant $g'(x)g'(y)=\frac{1}{c^2}(xy)^{\frac{1}{c}-1}\\ ,$ and since $x\ge L$ and $y\ge L$ in $B$ $$|A|=|(g\times g)(B)|=\int_{B} g'(x)g'(y)dxdy\le \left( \frac{L^{\frac{1}{c}-1}}{c}\right)^2|B|\\ . $$

Let's prove that the factor in front of $|B|$ is less than or equal to $1$. So we wish to show that $$L^{\frac{1}{c}-1}\le c\\ . $$ Recalling that $L=(1-L)^c$, that is, $c=\frac{\log L}{\log (1-L)}$, this reduces to

$$\frac{(1-L)\log(1- L)} {L\log L}\le 1 $$

which is indeed true exactly for $L\le 1/2$. So we have $$\int_0^1 F_{n+1}dx -\int_0^1 F_n dx = \int_0^L(F_{n+1}-F_n)dx + \int_L^1(F_{n+1}-F_n)dx=$$ $$= -|A|+|B|\ge 0\\, $$ as required.

Let $c>1$ be any fixed real exponent and let us denote, for any $x\in I:=[0,1]$, $f(x):=(1-x)^c$, and $g(x):=1-x^{\frac{1}{c}}$. So $f$ is a strictly decreasing homeomorphism of $I$ into itself, with inverse map $g$ and with the interior fixed point $0 < L< 1/2$. Also, since all even-order iterated of $f$ are strictly increasing, for any $x\in I$ and for any $n\in\mathbb{N}$, we have $$x\le L\quad\mathrm{iff}\quad f(x)\ge x\quad\mathrm{iff}\quad F_1(x)\le x \quad\mathrm{iff}\quad F_{n+1}(x)\le F_n(x)\\ .$$ The sets $$A:=\{(x,y)\in I^2\\ : \\ F_{n+1}(x)\le y \le F_n(x)\}$$ and $$B:=\{(x,y)\in I^2\\ : \\ F_{n+1}(x)\ge y \ge F_n(x)\}$$ are therefore included in the squares $[0,L]^2$, respectively $[L,1]^2$, so we have $$-\int_0^L(F_{n+1}-F_n)dx=|A|$$ and $$\int_L^1(F_{n+1}-F_n)dx=|B|\\ .$$ Moreover, $(x,y)\in A$, that is $ F_{n+1}(x)\le y \le F_n(x)$, if and only if $$F _ {n+1}(f(x))=f(F _ {n+1}(x))\ge f(y) \ge f(F_ n(x))=F_ n(f(x))\\ ,$$ that is, $(f\times f)(x,y):=(f(x), f(y))\in B$. So $A=(g\times g)(B)\\ .$ The map $g\times g$ has Jacobian determinant $g'(x)g'(y)=\frac{1}{c^2}(xy)^{\frac{1}{c}-1}\\ ,$ and since $x\ge L$ and $y\ge L$ in $B$ $$|A|=|(g\times g)(B)|=\int_{B} g'(x)g'(y)dxdy\le \left( \frac{L^{\frac{1}{c}-1}}{c}\right)^2|B|\\ . $$

Let's prove that the factor in front of $|B|$ is less than or equal to $1$. So we wish to show that $$L^{\frac{1}{c}-1}\le c\\ . $$ Recalling that $L=(1-L)^c$, that is, $c=\frac{\log L}{\log (1-L)}$, this reduces to

$$\frac{(1-L)\log(1- L)} {L\log L}\le 1 $$

which is indeed true exactly for $L\le 1/2$. So we have $$\int_0^1 F_{n+1}dx -\int_0^1 F_n dx = \int_0^L(F_{n+1}-F_n)dx + \int_L^1(F_{n+1}-F_n)dx=$$ $$= -|A|+|B|\ge 0\\, $$ as required.