real analysis - Is this proof that if $a_{n+1} = sqrt{2 + sqrt{a_n}}$ and $a_1 = sqrt{2}$, then $sqrt{2} leq a_n leq 2$ correct?

Let $a_1 = \sqrt{2}$ and $a_{n+1} = \sqrt{2 + \sqrt{a_n}}$. Now I want to show by induction that $\sqrt{2} \leq a_n \leq 2$ for all $n$.

The base case is $n=1$ and it is clear $\sqrt{2} \leq a_1 \leq 2$. Then I assume that $\sqrt{2} \leq a_n \leq 2$ holds and I want to show $\sqrt{2} \leq a_{n+1} \leq 2$.

Then I note that $\sqrt{2} \leq a_{n+1} \leq 2 \implies \sqrt{2} \leq \sqrt{2 + \sqrt{a_n}} \leq 2$. By squaring both sides I get $2 \leq 2 + \sqrt{a_n} \leq 4$. Then by subtracting $2$, I get $0 \leq \sqrt{a_n} \leq 2$. This means that $0 \leq a_n \leq 4$. This is ok because I assumed $0 \leq \sqrt{2} \leq a_n \leq 2 \leq 4$.

Edit
How about this: Since I know $\sqrt{2} \leq a_n \leq 2$. Then it is clear that $0 \leq a_n \leq 4$. By taking square roots I get $ 0 \leq \sqrt{a_n} \leq 2$. Now if I add 2, $2 \leq \sqrt{a_n} + 2 \leq 4$. Taking another square root I get $\sqrt{2} \leq \sqrt{\sqrt{a_n} + 2} \leq 2$. So $\sqrt{2} \leq a_{n+1} \leq 2$.

Answer

Note that you want to show the case $n$ implies $n+1$, but you're going the other way.

The good thing about this sequence is the inductive step is indeed easy.

The base case is $\sqrt 2 \leq a_1 \leq 2$

We now assume $\sqrt 2 \leq a_n \leq 2$

Take a square root, then add two, then take a square root.

$$\sqrt{\root 4 \of 2+2} \leq \sqrt{2+\sqrt a_n}\leq \sqrt{\sqrt2+2}$$

$$\sqrt{\root 4 \of 2+2} \leq a_{n+1}\leq \sqrt{\sqrt2+2}$$

Note that since $\sqrt 2 + 2 < 4 \Rightarrow \sqrt {\sqrt 2 + 2} < 2$ and similarily $2 < \root 4 \of 2 + 2 \Rightarrow \sqrt 2 < \sqrt {\root 4 \of 2 + 2} $

Add: Your edit makes perfect sense.