# How do I prove that f(x,y) <= g(x,y)

## Homework Statement

The problem is to prove that f(x,y) >= g(x,y)

x and y are positive integers

## Homework Equations

f(x,y) = 1/6 { 2(x-y-2)/(x+2y-2) + 10(x-y+1)/(x+2y+1) }

g(x,y) = 2(x-y)/(x+2y)

## The Attempt at a Solution

I'm looking for any general advice on how to approach this question. The approach that I have attempted is to set g(x,y) = k and then to solve for x.
x = -2y(k +1) / (k-2)

Then I substituted this formula back into the equation f(x,y) in the hopes that this would simply to a form that indicates that the inequality is true; however, after doing this a simple solution does not seem to fall out:

f(x,y) = 1/6 { (yk-2y-2k+2)/(-3y-k+2) + 10(-3yk-k-2)/(-6y+k-2))}

Any suggestions to how to approach this kind of problem or on this specific problem would be very appreciated!

Thanks
Dan

Last edited:

Päällikkö
Homework Helper
Well, I'd look for the minimum of f - g. Did this help?

Dear Päällikkö,

That does seem to be a good suggestion, if I come to a good solution I will post it

Thanks!
Dan

Dick
Homework Helper
I looks to me like it's actually pretty easy to solve f(x,y)=g(x,y) for a relation between x and y. That would give you the boundary curve separating the region where f(x,y)>g(x,y) from the region where f(x,y)<g(x,y).

Would you be willing to elaborate a bit more about how one could solve f(x,y)=g(x,y)?
Dan

Dick