$a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $ 
Let $a_n$ be a sequence satisfying $a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $

Attempt at solution using strong induction:
Base cases: $n = 1$ and $n = 2 \implies a_1 = a_2 = 1 \implies 1\le 1 \le 2$
Inductive assumption (strong induction): Assume that for all $m\in \mathbb{N}$ such that $1\le m \le k$, where $k\in \mathbb{N}$, the condition $1\le a_m \le 2$ holds True.
Show that $m = k+1$ holds true
$a_{k+1} = \frac{1}{2} \cdot (a_k + \frac{2}{a_{k-1}})$
I know that $a_k$ and $a_{k-1}$ are satisfying $ 1\le a_m\le2$ but I am not sure how to use that to prove that $a_{k+1}$ holds true for the condition.

Hint: Try maximizing and minimizing $ (a_k + \frac{2}{a_{k-1}})$ using 
(1) If $a \le c$ and $b \le d$ then $a + b \le c + d$
(2) If $a \ge c$ and $b \ge d$ then $a + b \ge c + d$
