Problem. Let $a_n$ be a sequence satisfying $$ a_1 = a_2 = 1 \qquad\text{and}\qquad a_n = \frac{1}{2}\left( a_{n-1} + \frac{2}{a_{n-2}} \right) $$ for $n \ge 3$. Prove that $ 1\le a_n \le 2 $ for all $n \in \mathbb{N}$.
What I did:
Base Case: If $n = k = 3$,
$$a_3 = \frac{1}{2} \left( a_{2} + \frac{2}{a_{1}} \right) = \frac{1}{2}\left(1 + \frac{2}{1}\right) = 1.5, $$
which is $\ge 1$ and $\le 2$. Therefore base case is true.
Inductive Step: Assume $1 \le a_n \le 2$ for $n = k \geq 3$. Consider $n = k + 1$.
$$ 1 \le a_{k+1} \le 2 \implies 1 \le \frac{1}{2}\left(a_{k} + \frac{2}{a_{k-1}}\right) \le 2$$
From here, I am stuck, unsure what to do next. I understand strong induction is needed here, but I don't know how exactly to implement it.
Apologies for poor $\LaTeX$ formatting, it's my first time.
Your question is a very good question. According to your question $a_n=\frac{1}{2}(a_n-1+\frac{2}{a_n-2})$ for $n\geq3$ . And $a_1=a_2=1$ . Now try to find $$\lim_{n\to\infty} a_n$$ . Now $$\lim_{n\to \infty} a_n$$ = $$\lim_{n\to \infty}\frac{1}{2}(a_n-1+\frac{2}{a_n-2})$$. So, $a_\infty=\frac{a_\infty}{2}+\frac{1}{a_\infty}$ . So, $(a_\infty)^{2}=2$ . So, $a_\infty=\sqrt{2}$ . Therefore $$\lim_{n\to \infty}a_n=\sqrt{2}$$ . Why I took $+\sqrt{2}$ and not $-\sqrt{2}$ because $a_n$ will lie between $[1,2]$ for every $n\geq3$ . So, your condition is proved i.e. $1\leq a_n\leq2$ .