<s>[INST] $a_{1}=a_{2}=1$ and $ a_{n}=\tfrac{1}{2} (a_{n-1}+\tfrac{2}{a_{n-2}}) $ for $n\geq 3$. Then for all $n\in \mathbb{N}$, $1\leq a_{n} \leq 2$ 
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*Let $a_{n}$ be the sequence satisfying $a_{1}=a_{2}=1$ and $ a_{n}=\tfrac{1}{2} (a_{n-1}+\tfrac{2}{a_{n-2}}) $ for $n\geq 3$. Then for all $n\in \mathbb{N}$, $1\leq a_{n} \leq 2$
For $n=3$, then we have $1\leq a_n\leq 2.$ 
Now assume for $n$ holds. Show for $n+1$ holds.
So
$$ a_{n+1}=\tfrac{1}{2} (a_{n}+\tfrac{2}{a_{n-1}}) $$
Then, what should I do now?
 [/INST] Assume $1\le a_k\le 2$ for all $k<n.$ Then $\frac12(a_{n-1}+\frac2{a_{n-2}})\leq \frac12(2+2/1)=2$  and similarly $\frac12(a_{n-1}+\frac2{a_{n-2}})\ge \frac12(1+2/2)=1$ because $1\leq a_k\leq 2$ for all $ k\le n-1.$
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