Let the sequence $\{a_n\}$ be defined as $a_1=1$ and $a_{n+1} = \frac{6a_n+3}{a_n+4}$. Show that $a_n \lt 3$ and the sequence is increasing. Let the sequence $\{a_n\}$ be defined as $a_1=1$ and $$a_{n+1} = \frac{6a_n+3}{a_n+4}$$
Then I'm asked to show :
$1)$ $a_n \lt 3$.
$2)$ Assuming $a_n \lt 3$, show that the sequence is increasing.
For the first part I tried to use Induction. When I assumed $a_n \lt 3$ and went on to prove that this implies $a_{n+1} \lt 3$, I got stuck :
$$ a_{n+1} = \frac{6a_n+3}{a_n+4} \lt \frac{6\cdot 3+3}{a_n+4} \lt \frac{21}{a_n+4}.$$
I'm stuck here can anyone please help?
Edit : I can do the second part easily. And also I got an appropriate answer by now for the first part.

Just for your curiosity.
If you look at this question, I described the steps for a first-order rational difference equation. If you just follow the steps, your coefficients being nice, you should end with
$$a_n=\frac{9\ 7^n-7\ 3^n}{7\ 3^n+3\ 7^n}=3-\frac{28\ 3^n}{7\ 3^n+3\ 7^n}$$
