Proving a simple inequality. 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.

Inequality is equivalent to the :
$4n^2+4n-1>0$
which is equivalent to the :
$(2n+1)^2>2$
You can use induction to prove this inequality :


*

*for $n=1$ it follows $9>2$

*suppose $(2n+1)^2>2$

*we want to to prove that $(2n+3)^2>2$ 
Since $(2n+1)^2>2$ it follows :
$4n^2+4n+1 > 2$
$4n^2+4n+1+8n+8>2+8n+8>2$
Hence :
$(2n+3)^2>2$
Q.E.D.
