I am trying to prove that
$\frac{n}{4n^2+1} > \frac{n+1}{4(n+1)^2+1}, \forall n\in\mathbb{N}$.
What I did so far was
$n < n+1\\ \Rightarrow \frac{n^2}{n} < \frac{(n+1)^2}{n+1}\\ \Rightarrow \frac{n}{n^2} > \frac{n+1}{(n+1)^2}\\ \Rightarrow \frac{n}{4n^2} > \frac{n+1}{4(n+1)^2}\\ \Rightarrow \frac{n}{4n^2+1} > \frac{n+1}{4(n+1)^2+1}\\ $
I'm not so sure about the last step though... basically, is
$\frac{a}{b} < \frac{c}{d} \Rightarrow \frac{a}{b+1} < \frac{c}{d+1} (b\neq -1 \wedge d \neq -1)$ a correct assumption? It was just a gut feeling for me, and I can't really justify it.
And yes, this is a homework question. I just didn't know what exactly I would have to look for, so I had to ask.