Q:

Convergence of $a_{n+2}^3+a_{n+2}=a_{n+1}+a_n$

Let $(a_n) _{n\ge 0}$ $a_{n+2}^3+a_{n+2}=a_{n+1}+a_n$,$\forall n\ge 1$, $a_0,a_1 \ge 1$. Prove that $(a_n) _{n\ge 0}$ is convergent.
I could prove that $a_n \ge 1$ by mathematical induction, but here I am stuck. 

A:

There might be an easier and more elegant solution, but this should work.
First observe that:
$$
a_{n+1}+a_{n}=a_{n+2}^3+a_{n+2}\geq2\sqrt{a_{n+2}^4}=2a_{n+2}^2\geq4a_{n+2}-2
$$
Here I used the AM-GM inequality and the simply fact that $(a_{n+2}-1)^2\geq0$.
Hence, $a_n\leq b_n$ with
$$
b_{n+2}=\frac{b_{n+1}+b_{n}+2}{4}\\
b_{0,1}=a_{0,1}
$$
Now, the recurrence equation for this reads:
$$
b_{n}=1 + (\tfrac{1 - \sqrt{17}}{8})^n C_1 + (\tfrac{1 + \sqrt{17}}{8})^n C_2
$$
Since $|\tfrac{1 \pm \sqrt{17}}{8}|<1$ we can deduct that $b_n\to1$ for $n\to\infty$. So together with $1\leq a_n\leq b_n$ we can conclude that $a_n\to1$.

