Show limit for recursive sequence by induction Let $(a_n)_{n \in \mathbb N}$ be a recursive sequence. It is defined as $a_1=1, \quad a_{n + 1} = \frac{4a_n}{3a_n+3}$. 

I have to show by induction, that it is limited for $a_n \ge \frac{1}{3}$.

So as an induction step I have $n \to n + 1$. And the idea is to get $a_{n+2}$ and show somehow $a_{n+2} \ge \frac{1}{3}$

$$a_{n+1} = \frac{4a_n}{3a_n + 3} $$
$$a_{n+2} = \frac{4 \frac{4a_n}{3a_n + 3}}{3 \frac{4a_n}{3a_n + 3} + 3} = \frac{ \frac{16a_n}{3a_n+3} }{ \frac{12 a_n}{3a_n + 3} + 3}$$

Question: How should I go on or am I on the wrong path?

Edit: You can find other answers for the exact same question here.
You wonder if $a_{n+1}\geq {\frac{1}{3}}$, so if

$$\frac{4a_n}{3a_n+3}\geq {\frac{1}{3}}$$ but this is equivalent to $$ 4a_n \geq a_n+1$$
or $$3a_n\geq 1$$
which is true by induction assumption.
To show that $a_{n+2}\ge1/3$, given that $a_{n+1}\ge1/3$, I would just plot the formula
$$
a_{n+2}=\frac{4a_{n+1}}{3a_{n+1}+3}
$$
as a function of $a_{n+1}$. This is clearly a strictly increasing function for $a_{n+1}\ge1/3$, which has limit $4/3$, and it takes on its minimum value ($=1/3$) at $a_{n+1}=1/3$. So $a_{n+2}\in[1/3,4/3)$. So your claim holds and the induction step from $n+1$ to $n+2$ is proved.