I have a recursive function defined as follows,
$$a_{1}=1,a_{k+1}=\sqrt{2a_{k}+3} \: \forall k\in \mathbb{N}$$
I think I found a very quick way to demonstrate that the function is increasing, but I want to know if it is valid as the proof techniques my professor has used for these sort of questions have been more drawn out and longer than this.
By induction, I will assume, $a_{k}\leq a_{k+1}$. The base case clearly holds.
Considering $a_{k+1}$, using our assumption and the definition of the function we have, $$\sqrt{2a_{k}+3}\leq \sqrt{2a_{k+1}+3}=a_{k+2}$$ This completes the proof. Is there any reason why this might not be a valid line of argument? If I really wanted to, I could construct the inequality as follows,
$$a_{k}\leq a_{k+1}\rightarrow2a_{k}\leq2a_{k+1}\rightarrow2a_{k}+3\leq2a_{k+1}+3\rightarrow\sqrt{2a_{k}+3}\leq\sqrt{2a_{k+1}+3}\rightarrow a_{k+1}\leq a_{k+2}$$
This is more explicit but still shorter than some of the proof techniques I've seen.
Yes, your proof is valid. Your first proof uses the fact that $x\mapsto \sqrt{2x+3}$ is an increasing function, which you prove at the end of your post using the fact that $x\mapsto \sqrt{x}$ is increasing. I suppose a complete proof would include showing that $x\mapsto \sqrt{x}$ is increasing.