For instance,

$\displaystyle a_{n+2}-a_{n+1}=\left(1+\frac{1}{a_{n+1}}\right)-\left(1+\frac{1}{a_{n}}\right)$ $\displaystyle =\frac{a_n-a_{n+1}}{a_na_{n+1}}$,

and $\displaystyle a_n a_{n+1}=a_n+1$ by the recurrence equation, hence if you know $\displaystyle 1\leq a_n\leq 2$, then $\displaystyle 2\leq a_na_{n+1}\leq 3$.

From this, you can conclude.

Sidenote: $\displaystyle a_n=1+\frac{1}{a_{n-1}}=1+\cfrac{1}{1+\cfrac{1}{a_{n-2}}}=\cdots=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+ \cdots+\cfrac{1}{1}}}}$ with $\displaystyle n-1$ fractions.