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Math Help - prove inequality

  1. #1
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    prove inequality

    x is less then or equal to 0
    y is less then or equal to 0
    then prove (xy)^1/2 less then or equal to x+y/2

    I was wondering if someone could tell me where to start It would be greatly appreiated i forgot this on my last post. Thank you very much. I am going crazy.
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  2. #2
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    This is the arithmetic-mean geometric mean inequality.

    Let x>=0 and y>=0.

    Then,
    0<=(x-y)^2 because a square is always non-negative.
    Expand,
    0<=x^2-2xy+y^2
    Add 4xy to both sides,
    4xy<=x^2+2xy+y^2=(x+y)^2
    Take square roots since both are non-negative,
    sqrt(4xy)<=sqrt(x+y)^2
    2sqrt(xy)<=|x+y|=x+y
    Divide by 2,
    sqrt(xy)<=(x+y)/2
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by schinb64 View Post
    x is less then or equal to 0
    y is less then or equal to 0
    then prove (xy)^1/2 less then or equal to x+y/2

    I was wondering if someone could tell me where to start It would be greatly appreiated i forgot this on my last post. Thank you very much. I am going crazy.
    as (xy)^(1/2) is positive, and (x+y)/2 is negative this is not true
    the way you have asked the question.

    ImPerfectHacker has answered this but for:

    x >= 0
    y >= 0,

    which is almost certainly what you were really asked to do.

    RonL
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  4. #4
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    Thumbs up Thank you for the responses you guys are great. The help is greatly appreciated

    Thank you for the help again. You guys are great. You have helped me so much thank you again
    Quote Originally Posted by ThePerfectHacker View Post
    This is the arithmetic-mean geometric mean inequality.

    Let x>=0 and y>=0.

    Then,
    0<=(x-y)^2 because a square is always non-negative.
    Expand,
    0<=x^2-2xy+y^2
    Add 4xy to both sides,
    4xy<=x^2+2xy+y^2=(x+y)^2
    Take square roots since both are non-negative,
    sqrt(4xy)<=sqrt(x+y)^2
    2sqrt(xy)<=|x+y|=x+y
    Divide by 2,
    sqrt(xy)<=(x+y)/2
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