Comparing two functions

1. Oct 1, 2012

the_kid

1. The problem statement, all variables and given/known data
I'm trying to show that the function f(x)=$\frac{1}{x^{2}}$ is greater than g(x)=$\frac{1}{x}$-ln(1+$\frac{1}{x}$) for all x>0.

2. Relevant equations

3. The attempt at a solution
This is really simple, but I'm having some trouble showing it rigorously. I thought about trying to compare the derivatives, but it's not that simple. Are there any good ways to do this?

2. Oct 1, 2012

SammyS

Staff Emeritus
Well... then the function (f-g)(x) must be positive ...

the function (f/g)(x) must be greater than 1, or be negative ...

the function (g/f)(x) must be less than 1, or be negative.

3. Oct 1, 2012

the_kid

Right. That was the other method I tried. I just can't figure out a way to do it without some "hand-waving."

4. Oct 2, 2012

jbunniii

Not sure if this simplifies anything, but if $0 < x < 1$, then $1/x^2 > 1/x$ so clearly $f(x) > g(x)$. So you only have to worry about $x > 1$.

5. Oct 2, 2012

HallsofIvy

Staff Emeritus
Using that, a fourth method: it is easy to show this is true for x= 1 so if you can prove that f'(x)> g'(x) for all x>1 you are done.

6. Oct 2, 2012

SammyS

Staff Emeritus
How about comparing f(1/x) and g(1/x) ?

7. Oct 13, 2012

the_kid

Hmmm, OK, how does this help me?

8. Oct 13, 2012

Ray Vickson

Well, derivatives are a bit simpler, etc.

RGV

9. Oct 14, 2012

the_kid

I've been able to show that f(1/x)>g(1/x) for all x>1. From this, how exactly can I conclude that f(x)>g(x)?

10. Oct 14, 2012

SammyS

Staff Emeritus
That shows that f(x)>g(x) for 0 < x < 1, which isn't a big help.