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Check my inequality proof please

  1. Sep 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove for all real numbers x and y that [tex]2xy =< x^2 + y^2[/tex]

    2. Relevant equations



    3. The attempt at a solution

    Well, since this is a problem regarding proof, I thought I would start with a contradictory statement like:

    2xy >= x^2+y^2



    0 >= x^2-2xy+y^2



    0 >= (x-y)^2

    Since (x-y)^2 is either a positive integer and a zero,

    0 =< (x-y)^2



    0 =< x^2-2xy+y^2



    2xy =< x^2+y^2

    Well that's all I can think of... can anyone point out any mistakes or anything? Is there more to it than just this?
     
  2. jcsd
  3. Sep 18, 2009 #2
    i think that it looks fine
     
  4. Sep 19, 2009 #3

    VietDao29

    User Avatar
    Homework Helper

    What you've done is not a Proof by Contradiction.

    The bottom part of your post is indeed qualified as a direct proof. No need for Contradiction. :)

    And btw, the negation of 2xy <= x2 + y2 is not

    Instead, it should be: 2xy > x2 + y2.
     
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