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Inequality proof

  1. Jun 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that if x and y are not both 0 then x2 + xy + y2 > 0


    2. Relevant equations



    3. The attempt at a solution
    Proof. Let x and y be arbitrary real numbers. Suppose x and y are not both zero. We will consider three cases.
    Case 1. x = y. Then x2 + xy + y2
    = x2 + x2 + x2
    = y2 + y2 + y2 = x2 + xy + y2 > 0 since the square of a real number is
    always greater than or equal to zero and x= y [tex]\neq[/tex] 0.
    Case 2. x > y. Then (x - y) > 0 and x3 - y3 > 0.
    Therefore, x3 - y3 = x3 + 0 - y3
    = x3 + (x2y - x2y + xy2 - xy2) - y3 = ( x - y)( x2 + xy + y2) > 0.
    Dividing both sides by ( x - y) we get x2 + xy + y2 which is greater
    than zero since x and y are not both zero.
    Case 3. x < y. Then y - x > 0 and y3 - x3 > 0. Therefore,
    y3 - x3 = y3 + 0 - x3
    = y3 + ((x2y - x2y + xy2 - xy2) - x3 = (y - x)( x2 + xy + y2) > 0.
    Dividing both sides by ( y - x) we get x2 + xy + y2 which is greater than zero since x and y are not both zero.

    Truth?
    I greatly appreciate anyone's help. I'm studying independently and don't have anyone to check my proofs
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 9, 2009 #2

    tiny-tim

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    Hi nike5! :smile:
    eugh! :yuck:

    you mustn't try to solve everything by the trichotomy law!

    Hint: complete the square :wink:
     
  4. Jun 9, 2009 #3
    ahhhh I feel like an idiot for not seeing that.

    Proof. Let x and y be arbitrary real numbers. Suppose x and y do not both equal 0.
    Then x2 + xy + y2 = (x2 + xy) + y2
    = ( x + (y / 2))2 + (3y2) / 4. Therefore,
    ( x + (y / 2))2 [tex]\geq[/tex] 0 and (3y2) / 4 [tex]\geq[/tex] 0 since the square of any real number is positive. Under the closure property of addition,
    ( x + (y / 2))2 + (3y2) / 4 [tex]\geq[/tex] 0 and since x and y are not both zero we can conclude that ( x + (y / 2))2 + (3y2) / 4
    = x2 + xy + y2 > 0.
     
  5. Jun 9, 2009 #4

    tiny-tim

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    :biggrin: Woohoo! :biggrin:
     
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