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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

3 Answers 3