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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. –  Mark Bennet Apr 8 '12 at 13:02
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3 Answers

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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Ok, then my thinking was wrong, but the solution in this case correct? Thank you very much. Oh you just removed the second part. So was that invalid? –  Cubic Apr 8 '12 at 12:49
    
Yes, but it's fine now. –  Barry Smith Apr 8 '12 at 12:52
    
@Cubic If you have classical logic at work, then the reasoning you gave does work out as correct and you've given a proof, since the last two inequalities I1, I2 you have written are true, which implies (I1->I2) as true. However, this probably doesn't help you for your course. –  Doug Spoonwood Apr 8 '12 at 13:07
    
Oh, I see what you did there. –  Cubic Apr 8 '12 at 13:08
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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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Isn't $4n^2 +4n -1 > 0$ already enough, given that $n\in\mathbb{N}$? –  Cubic Apr 8 '12 at 13:13
    
@Cubic You can't just say "it is obvious" , you have to prove it... –  pedja Apr 8 '12 at 13:16
    
That's the point of the last "clause" of my answer. Your teacher should give you some guidance as to acceptable first principles from which you can work (i.e., which statements are "obvious"). An inductive proof would be overkill if I were teaching this course (unless we were studying how to use induction). The natural numbers are closed under multiplication and addition and are all at least $1$. $4$ and $n^2+n$ are both natural numbers, so $4n^2+4n$ is at least $1$. It could be said that it is clearly bigger than $1$. Or if your teacher is super picky, say $4>1$, so $4n^2+4n>n^2+n>1+1>1$ –  Barry Smith Apr 8 '12 at 13:28
    
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. –  Cubic Apr 8 '12 at 16:23
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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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Or better yet, present both types of solution! –  Barry Smith Apr 8 '12 at 14:16
    
That's supposed to be the proof for the monotocity of a series, so using the monotocity of a function for that would be kinda weird. –  Cubic Apr 8 '12 at 16:16
1  
I think, by "series", you mean sequence. That's only weird if you haven't seen the formal definition of a sequence: a function whose domain is the natural numbers (or a variant of this). The definition of "increasing function" is the same whether the domain is a set of real numbers or the natural numbers, and so you can translate statements about functions on the real numbers to statements about their restrictions to the natural numbers. –  Barry Smith Apr 8 '12 at 19:55
    
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 $x<y$, then surely $f(n)>f(n+1)$. –  AD. Apr 8 '12 at 20:03
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