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.

If $a_n \lt 3$, then
$$a_{n+1}= \frac {6a_n+3}{a_n+4}=\frac{6a_n+24-21}{a_n+4}=6-\frac{21}{a_n+4} \lt 6-\frac{21}{7}=3.$$
