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Quadratic inequality, proof.

  1. Mar 3, 2012 #1
    1. The problem statement, all variables and given/known data

    By expanding (x-y)^2, prove that x^2 +y^2 ≥ 2xy for all real numbers x & y.


    2. Relevant equations



    3. The attempt at a solution
    expanding (x-y)^2

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

    Hence, x^2 + y^2 = 2xy

    But where does the ≥ come into it? and why?
    when you put values in (except i which is not real of course) they all come out as = 2xy, which does satisfy ≥2xy, but why does this come into it??
    Some insight would be fantastic!
     
  2. jcsd
  3. Mar 3, 2012 #2

    Dick

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    If x and y are real then (x-y) is real. (x-y)^2>=0, yes?
     
  4. Mar 3, 2012 #3
    Ohhhh I see...
    Thanks!
     
  5. Mar 4, 2012 #4
    Continuing from Dick's hint:

    (x-y)^2 ≥ 0 (Trivial inequality)
    x^2-2xy+y^2 ≥ 0
    And adding 2xy to both sides, we get:
    x^2+y^2 ≥ 2xy as desired.

    And if you're up for it, try proving the two variable case of the AM-GM inequality from here (it's pretty simple).
     
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