Convergence of sequence $a(n+1) = \frac{3(1+a(n))}{3+a(n)}$

The sequence is defined by:

$$a(1)=1, a(n+1) = \frac{3(1+a(n))}{3+a(n)}$$

How do I show that the sequence is monotonic and bounded from above?

My approach to show its monotone was:

$$a(n+1) - a(n) \geq 0$$

$$\frac{3(1+a(n))}{3+a(n)} - a(n) = \frac{3+3a(n)}{3+a(n)} - \frac{3a(n)+ a(n)^2}{3+a(n)} = \frac{3(1+a(n)) - a(n)(3+a(n))}{3+a(n)} = \frac{3-a(n)^2}{3+a(n)} \geq ? \geq 0$$

I know the sequence is bounded from above by $$\sqrt{3}$$ but I'm totally lost at how to show this.

Use induction to show that $$a(n) <\sqrt 3$$ for all $$n$$. Then use the fact that $$\frac {3(1+x)} {3+x} >x$$ for $$0 to show that $$a(n+1) >a(n)$$ for all $$n$$. You can also check that $$a(n) <2$$ for all $$n$$ so $$(a(n))$$ is convergent.
I will let you show that the limit is $$1$$.
[It is useful to start with the observation that $$\frac {3(1+x)} {3+x}$$ is an increasing function on $$(0,\infty)$$].
Notice that $$\dfrac{a_{n+1}-\sqrt{3}}{a_{n+1}+\sqrt{3}}=\dfrac{a_{n}-\sqrt{3}}{a_{n}+\sqrt{3}}\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$$. We can calculate the accurate value of $$a_n$$.