<s>[INST] How to prove this inequality for $a_{n}<5$ Let $a_{1}=1,a_{2}=2$ and be related by the recurrence
$$a_{n+2}=\dfrac{a^2_{n+1}(1+a_{n})+(1+2a_{n})a_{n+1}-a^2_{n}}{a_{n+1}+a^2_{n}+a_{n}+1}$$
for $n\in N.$
How can I show that
$a_{n}<5,n\in N$?
my idea: I think we have 
$$a_{n+2}=pa_{n+1}+qa_{n}$$
because we have
$$a_{n+2}a_{n+1}+a_{n+2}a^2_{n}+a_{n+2}a_{n}+a_{n+2}=a^2_{n+1}+a^2_{n+1}a_{n}+a_{n+1}+2a_{n}a_{n+1}-a^2_{n}\cdots (1)$$
then 
$$a_{n+1}a_{n}+a_{n+1}a^2_{n-1}+a_{n+1}a_{n-1}+a_{n+1}=a^2_{n}+a^2_{n}a_{n-1}+a_{n}+2a_{n-1}a_{n}-a^2_{n-1}\cdots (2)$$
(1)+(2)
 [/INST] Let us start with @Did's simplified form of the recurrence relation (with typos corrected): $$b_{n+2}=\dfrac{b_nb_{n+1}^2}{b_n^2-b_n+b_{n+1}}\quad (n=1,2,...)\quad\text{with}\quad b_1=2\text{ and }b_2=3.$$Write $c_n:=\dfrac {b_{n+1}+b_n^2}{b_nb_{n+1}}\;(n=1,2,...).$ Then substituting for $b_{n+2}$ in the corresponding expression for $c_{n+1}$ and simplifying gives$$c_{n+1}=c_n\quad (n=1,2,...).$$Therefore $c_n=c_1=\frac76$ forall $n$; that is $(b_{n+1}+b_n^2)/b_nb_{n+1}=\frac76,$ or$$b_{n+1}=\dfrac{6b_n^2}{7b_n-6}.$$Now the map$f:x\mapsto 6x^2/(7x-6)$ satisfies $f(1)=f(6)=6$ and $1<f(x)<6$ for $1<x<6$ (the minimum is $\frac{144}{49}$ at $x=\frac{12}{7}$). Since $b_1\;(=2)$ is in this range, so are all $b_n$. Thus $b_n<6$ for all $n.$
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