$a_1 = a_2 = 1$ 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.
 A: Assume $x,y\in [1,2].$ Then $x\le 2$ and $y\ge 1\implies \dfrac{1}{y}\le 1\implies  \dfrac{2}{y}\le 2.$ Thus
$$\dfrac12\left(x+\dfrac2y\right)\le \dfrac12(2+2)=2.$$ On the other hand $x\ge 1$ and $1\le y\le 2\implies \dfrac{1}{y}\ge\dfrac12 \implies  \dfrac{2}{y}\ge 1.$ Thus
$$\dfrac12\left(x+\dfrac2y\right)\ge \dfrac12(1+1)=1.$$
A: $$a_{n+1} = \frac{1}{2}a_{n} + \frac1{a_{n-1}}$$
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 $1 \leq a_{n+1} \leq 2$.
A: 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*}
A: You can find maximum and minimum values for $a_{k+1}$ by maximizing and minimizing summands respectively. That is:
$$1=\frac{1}{2}\Big(1+\frac{2}{2}\Big)≤a_{k+1}≤\frac{1}{2}\Big(2+\frac{2}{1}\Big)=2$$
A: 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 $a + b \ge c + d$
