<s>[INST] $a_1 = a_2 = 1$ 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.
 [/INST] $$a_{n+1} = \frac{1}{2}a_{n} + \frac1{a_{n-1}}$$
Assume that for all $k \leq n$ we have $1 \leq a_k \leq 2$. Under those constraints it's easy to see that the above formula for $a_{n+1}$ takes on maximal value of $\frac{1}{2}\cdot 2 + \frac{1}{1} = 2$ and minimal value of $\frac{1}{2} \cdot 1 + \frac{1}{2} = 1$. But both of these extremes still maintain $1\leq a_{n+1} \leq 2$.
Thus when for all $k \leq n$ we have $1 \leq a_k \leq 2$ we also have $1 \leq a_{n+1} \leq 2$.
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