The task is to prove that the iterations of the logistic map $F_r(x)=rx(1-x)$ for $2<r<3$ and $x \in \langle0,1\rangle$ converges to $x_r:=\frac{r-1}{r}$, in other words $$\lim_n F_r^{(n)}(x)=x_r$$.
Proof: Let $2<r<3$, then $F_r(\frac{1}{2})=\frac{r}{2}>\frac{1}{2}$, and $x_r=\frac{r-1}{r}>\frac{1}{2}$. Let $\hat x_r$ be the unique point in $\langle 0,\frac{1}{2}\rangle$ such that $F_r(\hat x_r)=x_r$ (this point is exactly $\frac{1}{r}$, but that is unimportant). By simple calculation $F^{(2)}_r(\frac{1}{2})=F_r(\frac{r}{4})=\frac{r^2}{16}(4-r)>\frac{1}{2}$. So since $F_r$ is decreasing on $\langle 0,\frac{1}{2}\rangle$ and increasing on $\langle\frac{1}{2},1\rangle$, it follows: $$F_r^{(2)}([ \hat x_r,x_r]) \subseteq F_r^{(2)}([ \hat x_r,\frac{1}{2}]) \cup F_r^{(2)}([\frac{1}{2},x_r])\subseteq F_r([\frac{1}{2},x_r]) \cup F_r ([\frac{1}{2},x_r])\subseteq [\frac{1}{2},x_r]$$.
This part of the proof I understood, but now the text says for $x \in [\frac{1}{2},x_r\rangle$, the sequence $(F_r^{(2n)}(x))_n$ increases and $(F_r^{(2n+1)}(x))_n$ decreases. I tried to prove it by solving the inequality $F_r^{2}(x)>x$ which would solve the proof,but i can't seem to solve that inequality (since $F_r^{2}$ is a degree $4$ polynomial). Is there a smart way to do this?