Proof by induction: Let $a_0=3$ and $a_{n+1}=\sqrt{a_n+7}$ if $n>0$, Prove: $3<a_n<4$

Question

Let $a_0=3$ and $a_{n+1}=\sqrt{a_n+7}$ if $n>0$

Prove: $3<a_n<4$

At first I was quite surprised it's actually true for the base cases:

$n=0$, $a_1=\sqrt{3+7}=\sqrt{10}$

$n=1$, $a_2=\sqrt{\sqrt{10}+7}$

I'm not sure here how to even construct the induction hypothesis, never had an exercise where I needed to prove that some variable is in some range.

*Taken out of a book called "A Walkthrough Combinatorics" and unfortunately, there's no solution for this exercise.

I'm not a student yet so I don't know any university level math.

Answer 1

Since $a_0=3$, $a_0+7=10$ and so $9<a_0+7<16$. Therefore, $3<a_1<4$.

Now, let $n\in\mathbb N$ and suppose that $3<a_n<4$. Then $10<a_n+7<11$ and therefore $9<a_n+7<16$. So, $3<a_{n+1}<4$.