Problem with induction of $a_{n+1} = (a_n^2 + 12)/7$

Question

sorry new user so sorry if this question is inappropriate or badly formatted.

I have the equation

$$a_{n+1} = \frac{{a_n}^2 + 12}{7} \, , \qquad n\geq1$$

where $3 < x < 4$. Prove by induction that $3 < a_n < 4$ for all $n$.

I know that $a_1 = x$, so $P(1)$ is true.

Next I assumed that $P(k)$ holds true so that $3 <a_k < 4$.

Next I had to prove $P(k+1)$ holds true and I had some issues here.

Here's what I have.

$$7a_{k+1} = {a_k}^2 + 12$$

I then assumed I had to sub $4$ in place of $a_k$ due to $P(k)$ so this gives

$$< 16 + 12 = 28$$

I then did this same process for $3$.

$$7a_{k+1} = {a_k}^2 + 12 = 21$$

Now what I have is apparently sufficient proof to prove the induction but I don't understand how so I must be misunderstanding the question. I thought that to prove true both numbers would have to be between $3$ and $4$ as

$$3 <a_{k+1} < 4$$

Thanks for any help!

Answer 1

If $3<a_n<4$, then $9<a_n^2<16$. Hence $21<a_n^2+12<28$, therefore (divide by $7$):

$$3<a_{n+1}<4$$

Answer 2

Hint : if $x\in [3,4]$, then $\frac{x^2+12}{7}\in[\frac{3^2+12}{7},\frac{4^2+12}{7}]$ (as you more or less noticed). Apply this to $x=a_k$.