let $ U_n $ the sequence defined as :$$ U_{n+1}= \dfrac{U^2_{n}+2}{U^2_{n}+1}$$ , With $U_{0}=1$ , I want to show for every natural number $n$ the following inequality :
$$ 2-U_{n+1} \leq \frac45 (2-U_n)$$ , Now my attempt , I have showed in the first step that :$ 0< U_n < 2 $ by induction and in the same time i have showed that $ U_n $ is a deacreasing sequence , Now i have used induction proof to show that $$ 2-U_{n+2} \leq \frac45 (2-U_{n+1})$$ , use the fact by assuming :$ 2-U_{n+1} \leq \frac45 (2-U_n)$, $n\in \mathbb{N}$ , Now we can use monotonicity of $ U_n $ since it is deacreasing we have :$$ 2-U_{n+1} \leq 2-U_{n+2},n\in \mathbb{N} $$ , Then we have the two following results:
$$ 2-U_{n+1} \leq 2-U_{n+2},n\in \mathbb{N}\tag{1} $$ and $$ 2-U_{n+1} \leq \frac45(2-U_{n}),n\in \mathbb{N}\tag{2} $$
From the two above results i can't deduce wether :$$ 2-U_{n+1} \leq \frac45 (2-U_n),n\in \mathbb{N}\tag{3} $$ , So where is my problem ? any help
The problem is what you want to prove is false. Given $f(x)=\frac{x^2+2}{x^2+1}$, $f(2)\ne 2$. Since $2$ is not a fixed point, the claim is false.
By the way if the sequence was defined as follow: $$u_{n+1}=\frac{u_n^2+2}{u_n^2-1},$$then you should be able to prove it.