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Math Help - more inequalities

  1. #1
    Senior Member furor celtica's Avatar
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    Wink more inequalities

    show that x^2 - 4xy + 5y^2 >= 0 for all real values of x and y.
    to do this must i simply prove that the quardatic has no real roots? does the absence of real roots always prove that all values of the given variable fulfill the conditions?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    x^2-4xy+5y^2
    =x^2-4xy+4y^2+y^2
    =(x-2y)^2+y^2\geq0
    (being the sum of 2 squares)
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  3. #3
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    Hello, furor celtica!

    Show that: . x^2 - 4xy + 5y^2 \:\geq\:0 for all real values of x and y.

    We have: . x^2 - 4xy + 4y^2 + y^2 \:=\:(x -2y)^2 + y^2

    The square of any real number is nonnegative.
    And the sum of two nonnegative numbers is nonnegative.

    Therefore: . (x-2y)^2 + y^2 \:\geq \:0

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  4. #4
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    the absence of real roots indicates that x^2 - 4xy + 5y^2 > 0 not x^2 - 4xy + 5y^2 >= 0 because if it does = 0 then there are roots. Btw I might be wrong
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  5. #5
    Senior Member furor celtica's Avatar
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    can somebody verify this?
    that was actually my main question, i know that it can be proved using perfect squares but i was wondering abouit the root part.
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