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 $a_n\to1$.
A: Initial remarks : 
a) In case of convergence to a limit $L$, we would have $L^3+L=L+L$, with solutions $L=-1,0,1$. 
b) We assume that, up to a switching operation, $a_1 \leq a_0.$

We are going to show that $a_n$ is convergent with limit $L=1.$

First step : the recurrence "definition" 
$$a_{n+2}^3+a_{n+2}=a_{n+1}+a_n$$
of sequence $a_n$, rewritten under the form
$$ f(a_{n+2})=a_{n+1}+a_n \ \ \text{where} \ \ f(x):=x^3+x, \tag{0}$$
cannot be considered as a "definition" unless it has been proved
that $a_{n+2}$ is determined in a unique way by (1). This will be the consequence of 
Lemma $0$ : $f$ is a bijection.
Proof : $f'(x)=3x^2+1\geq 1>0$. Thus function $f$ is strictly increasing, therefore is bijective. $\square$.
Let us write (0) under the form :
$$a_{n+2}=f^{-1}(a_{n+1}+a_n)\tag{1}$$
Lemma $1$ : $a_n \geq 1$ whatever $n$.
Proof : by recurrence, using the fact that $f^{-1}(2,+\infty)=(1,\infty). \ \square$
Lemma $2$ : for any $x \geq 0, f(x) \geq 2x^2$
Proof by observing that $x^3+x-2x^2=x(x-1)^2 \geq 0. \ \square$
As a consequence of lemma $2$, 
$$f^{-1}(u) \leq \sqrt{\tfrac{u}{2}}.\tag{2}$$
Thus, for any $n$, 
$$a_{n+2} \leq \sqrt{\tfrac{a_{n+1}+a_n}{2}}\tag{3}.$$
Let us define an auxiliary sequence $b_n$ by 
$$b_n=a_n-1\tag{4}$$ 
Our objective is thus to prove that $b_n \to 0$. (3) becomes :
$$b_{n+2} \leq \sqrt{1+\tfrac{b_{n+1}+b_n}{2}}-1\tag{5}.$$
Lemma $3$ : $b_n$ is a positive decreasing sequence bounded from below by $0$.
Proof : By recurrence. True (see initial remark b)) for the two first elements. For the general case, use in (5) the (classical) result : 
$$\text{for all} \ x>0, \ \ \ 1 \leq \sqrt{1+x}\leq 1+\tfrac{x}{2}.\tag{5}$$ 
$\ \square$
Lemma $3$ allows to conclude that $a_n$ is a positive decreasing sequence bounded from below by $1$. Thus $a_n$ converges to a limit which is necessarily $L=1$ (see initial remark a)).
Remark about the rate of convergence : I have observed that sequence $b_n$ behaves asymptotically as a geometrical sequence with ratio $r=0.640388...$ whatever the initial values $a_0$ and $a_1$. This value is in fact equal to the value $(1+\sqrt{17})/8$ found for one of the roots of characteristic equation found by @maxmilgram. I have no true proof of this fact.
A: This is a variant on maxmilgram's answer. If we write $a_n=1+u_n$ with $u_n\ge0$ (as the OP observes), the recursion becomes
$$u_{n+2}^3+3u_{n+2}^2+4u_{n+2}=u_{n+1}+u_n$$
This implies $4u_{n+2}\le u_{n+1}+u_n$, at which point we have $u_n\le v_n$ with $v_0=u_0$, $v_1=u_1$, and
$$v_{n+2}={v_{n+1}+v_n\over4}$$
so that
$$0\le u_n\le v_n=C_1\left(1-\sqrt{17}\over8\right)^n+C_2\left(1+\sqrt{17}\over8\right)^n\to0$$
as in maxmilgram's answer.
