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Homework Help: How do I prove that f(x,y) <= g(x,y)

  1. Mar 2, 2010 #1
    1. The problem statement, all variables and given/known data

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

    x and y are positive integers

    2. Relevant 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)

    3. 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!

    Last edited: Mar 2, 2010
  2. jcsd
  3. Mar 2, 2010 #2


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    Well, I'd look for the minimum of f - g. Did this help?
  4. Mar 2, 2010 #3
    Dear Päällikkö,

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

  5. Mar 2, 2010 #4


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    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).
  6. Mar 2, 2010 #5
    Would you be willing to elaborate a bit more about how one could solve f(x,y)=g(x,y)?
    Thanks in advance.
  7. Mar 2, 2010 #6


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    Actually, I cheated. I set f(x,y)=g(x,y) in a computer program and said 'solve for y'. The answer came out pretty simple. If you are doing it by hand I guess you'd have to clear out all the fractions by multiplying both sides by (x+2y-2)*(x+2y+1)*(x+2y), expand everything start cancelling. Offhand, I don't see any clever way to avoid that. Unless maybe substuting u=2x+y and v=x-y makes it easier to keep all the terms straight.
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