Proof of the sequence $a_{n+1}=\frac{4a_n}{3a_n+3}$ is limited.

Question

Let $(a_n)_{n\in \mathbb{N}}$ be a sequence with $a_1=1$ and $a_{n+1}=\dfrac{4a_n}{3a_n+3}$.

To show: $a_n\geq 1/3$

Proof by induction:

Base case: n=1 implies $1 \geq 1/3$

induction hypothesis: $a_n \geq 1/3 \forall n \in \mathbb{N}$

induction step: $n \mapsto n+1$

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

Now I cannot go on, how can I show by that $a_{n+1} \geq a_n \geq 1/3$? Thx

Answer 1

Observe$$a_{n+1}= \frac{4a_n}{3a_n+3}=\frac43-\frac4{3a_n+3},$$ where $a_n\ge\frac13$ implies $3a_n+3\ge 4$ and so $\frac4{3a_n+3}\le 1$ and henec $a_{n+1}\ge\frac13$.

Answer 2

You have: $a_{n+1} - \dfrac{1}{3} = \dfrac{4a_n}{3a_n+3} - \dfrac{1}{3}= \dfrac{3a_n-1}{3(a_n+1)}\ge 0$ since $a_n \ge \dfrac{1}{3}$. Thus $a_{n+1} \ge 1/3$ for all $n \ge 1$.