logic - $a_1 = a_2 = 1$ 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.

Answer

We have $a_k \geq 1$ and $a_{k-1} \leq 2$ ; which gives $\frac{2}{a_{k-1}} \geq 1$. So

$$\begin{eqnarray*} a_{k+1} = \frac{1}{2} \left( a_k +\frac{2}{a_{k-1}} \right) \geq \frac{1}{2} \left( 1+1 \right)=1. \end{eqnarray*}$$
We have $a_k \leq 2$ and $a_{k-1} \geq 1$ ; which gives $\frac{2}{a_{k-1}} \leq 2$. So
$$\begin{eqnarray*} a_{k+1} = \frac{1}{2} \left( a_k +\frac{2}{a_{k-1}} \right) \leq \frac{1}{2} \left( 2+2 \right)=2. \end{eqnarray*}$$