Show that the sequence defined recursively by $a_{n+1} = \sqrt{3a_n + 1}$ is increasing

Question

Assume that the sequence ${a_n}$ is defined recursively by $a_{n+1} = \sqrt{3a_n + 1}$ for all $n \in \mathbb N$, with $a_1 = 1$. Use mathematical induction to prove that $a_n \leq a_{n+1}$ for all $n \in \mathbb N$.

I've gotten most of the way, but I need help with the last bit. I've proven the base case, and gotten as far as:

Assume $P(k)$ is true. That is, $a_k \leq a_{k+1}$ for any $k \in \mathbb N$. Prove $P(k+1)$. That is, $a_{k+1} \leq a_{k+1+1}$.

And now I'm stuck.

Answer 1

$a_{n+1} - a_n = \sqrt{3a_n+1} - \sqrt{3a_{n-1}+1} = \dfrac{3(a_n-a_{n-1})}{\sqrt{3a_n+1}+\sqrt{3a_{n-1}+1}} \geq 0$ since by the inductive step: $a_n \geq a_{n-1}$, and the proof is completed.

Answer 2

Any time you have $a_{n+1}=f(a_n)$ where $f$ is an increasing function, the resulting sequence is monotone. Indeed, whichever inequality you have between $a_1$ and $a_2$ will propagate to subsequent terms:

$$a_{k+1}\ge a_k \implies f(a_{k+1})\ge f(a_k)$$ or $$a_{k+1}\le a_k \implies f(a_{k+1})\le f(a_k)$$

Answer 3

Assuming $a_n\geq a_{n-1}$ we have $3a_n\geq 3a_{n-1}$ and also $3a_n+1\geq 3a_{n-1}+1$. Now, since square root is a monotonic function, we get $\sqrt{3a_n+1}\geq\sqrt{3a_{n-1}+1}$. Eventually: $$a_{n+1}=\sqrt{3a_n+1}\geq\sqrt{3a_{n-1}+1}=a_{n} $$