Given $x_1 := \sqrt{2}$ and $x_{n+1} :=\sqrt{2x_n} $, prove $\sqrt{2} ≤ x_n ≤ 2$ 

Let $x_1 := \sqrt{2}$ and $x_{n+1} :=\sqrt{2x_n} $ for all $n \in \mathbb{N}$.
By using proof by induction:
(i) Prove that $\sqrt{2} ≤ x_n ≤ 2$ for all $n \in \mathbb{N}$.
(ii) Prove that $x_n ≤ x_{n+1}$ for all $n \in \mathbb{N}$.


For (i)
Let's prove the case base case, for $n=1$ we have that $\sqrt{2} \leq x_1\leq 2$ which is clearly true by $x_1=\sqrt{2}$ 
Now we assume true for $n=k$ $\Longrightarrow$ $\sqrt{2}\leq x_k\leq2$ and we are required to prove $\sqrt{2}\leq x_{k+1} \leq 2$.
From our hypothesis : $\sqrt{2}\leq x_k\leq2$ however we know that $x_{k+1} =\sqrt{2x_k}\Longrightarrow x_k = \frac{(x_{k+1})^2}{2}$ so we have  $\sqrt{2}\leq \frac{(x_{k+1})^2}{2}\leq2 \Leftrightarrow \sqrt{2\sqrt{2}}\leq x_{k+1} \leq 2$. 
How do I proceed?

Hints:


*

*$\;\displaystyle \frac{x_{n+1}}{2}=\sqrt{\frac{x_{n}}{2}}\,$, so $\;\displaystyle \frac{x_{n+1}}{2} \le 1 \;\iff\; \frac{x_{n}}{2} \le 1 \;\iff\;  \cdots  \;\iff\; \frac{x_1}{2} \le 1\,$

*$\;\displaystyle \frac{x_{n+1}}{x_n}=\sqrt{\frac{x_{n}}{x_{n-1}}}\,$, so $\;\displaystyle \frac{x_{n+1}}{x_n} \ge 1 \;\iff\; \frac{x_{n}}{x_{n-1}} \ge 1 \;\iff\;  \cdots  \;\iff\; \frac{x_2}{x_1} \ge 1\,$
