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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.

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    Note that the denominators are positive, so you can multiply through and get a cubic expression on each side. It should be obvious that the cubic terms on each side cancel. Then take all the terms to one side and get a quadratic an^2+2bn+c>0. Completing the square gives a(n + \frac b a)^2 +c-\frac {b^2}{a^2}>0. It is then easy to identify when the inequality holds. This is different from the method you have chosen to try, but brute force can sometimes be quicker.2012-04-08

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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.

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    Oh, I see what you did there.2012-04-08
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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 :

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

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

  3. 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.

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    Well, my homeworks are corrected by the assistants of my prof, but I'm pretty sure that what they're more interested in is reducing the problem to one that has a reasonably obvious solution. If I was to prove 'everything' I would have to go back to the peano axioms every time I wanted to add 2 natural numbers. I don't think that's in anyones interest here.2012-04-08
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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 $x_0>0$?

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    Cubic, if what Barry said sounds abstract, it is sufficient to note that the expression of yours is $f(n)$. Hence if you know that f(x)>f(y) for x, then surely f(n)>f(n+1).2012-04-08