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.
 A: No, the assumption you suggest is not always correct.  For instance, let $a=2$, $b=7$, $c=1$, and $d=3$.  Then 
$\frac{a}{b} < \frac{c}{d}$ is true, but
$\frac{a}{b+1} < \frac{c}{d+1}$ is not.
If the argument for such an inequality doesn't immediately jump to mind, do some scratch work and work backwards from the inequality you're trying to prove to something you know to be true.  Then reverse your argument and make sure each step was reversible.  Here, you should be able to work backwards to obtain 
$4n^2+4n > 1$
which is (probably) acceptable as a clearly true starting point for your argument since the left side is 4 times a natural number.
A: 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.
A: Since it is a homework in analysis, you might wish to study monotonicity of the function 
$$f(x)=\frac{x}{4x^2 +1}$$
In other words, can you find $x_0$ so that $f'(x)$ is negative for $x>x_0$? 
What is the smallest $x_0>0$?  
