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Homework Help: Another proof

  1. Sep 18, 2009 #1
    Hi everyone. I know i keep posting all these questions, but each question in my textbook just keeps on bringing on new challenges. Is there a rule against posting to many questions?

    1. The problem statement, all variables and given/known data

    \text{Prove that if}
    \text{are not both 0, then}

    2. Relevant equations


    3. The attempt at a solution

    This is the second part of a question I posted earlier (https://www.physicsforums.com/showthread.php?t=338240). I'm guessing the questions are related somehow, but all the methods used on the earlier question don't seem to work on this question.

    I tried grouping all the positive terms ([tex]x^4, y^4, x^2y^2[/tex]) and separating the equation based on those, and it works for the cases where x and y are both positive or both negative, but when they have opposite signs, it's impossible to figure out.

    Any hints? I know that multiplying by (x-y) gives [tex]x^5-y^5[/tex], but I don't know how much that helps.
    Last edited: Sep 18, 2009
  2. jcsd
  3. Sep 18, 2009 #2
    Iis the last term in the inequality supposed to be y2 or y4?

  4. Sep 18, 2009 #3
    ah yes, thank you, it's been fixed
  5. Sep 18, 2009 #4
    it's just the same as the previous part isn't it?
  6. Sep 18, 2009 #5
    well, i've tried this:

    x^4+x^3y+x^2y^2+xy^3+y^4 &> 0\\
    x^4+y^4+xy(x^2+xy+y^2) &> 0

    and we know from the previous part that x^2 + xy + y^2 is positive, but there's that xy term which doesn't make it as clear cut as the last question.
  7. Sep 18, 2009 #6
    (x^5 - y^5) / (x-y) ?

    btw, check your inbox if you haven't
  8. Sep 18, 2009 #7
    Consider cases where x > y and x < y.

    (The case where x = y is more easily dealt with in the original polynomial form.)

  9. Sep 18, 2009 #8
    yes, that's why I was confused - he came up with the same proof in his last thread
  10. Sep 18, 2009 #9
    haha, i guess come up with that in my last thread. oops.
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