Let $(a_n)_{n\in\mathbb{N}}$ satisfy that $a_1=2$ and $a_{n+1}=2+\frac{1}{a_n}$. Show that for all $n\in\mathbb{N}$ with $n\geq 2$, $$ |a_{n+1}-a_n|\leq\frac14|a_{n-1}-a_n| $$
So I can show the base case to be true, but I can't see how to show the inductive step.
$$\forall i\in \mathbb{N} \qquad a_{i}=2+\frac{1}{a_{i-1}}\geq2$$ $$a_{n+1}=2+\frac{1}{a_n}...(1)$$ $$a_{n}=2+\frac{1}{a_{n-1}}...(2)$$
Subtracting (2) fom (1), $$ |a_n-a_{n-1}|=|\frac{a_n-a_{n-1}}{a_na_{n-1}}|=\frac{|a_n-a_{n-1}|}{|a_na_{n-1|}}\leq\frac{|a_n-a_{n-1}|}{2\times 2}=\frac{|a_n-a_{n-1}|}{4}$$