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Math Help - algebra inequality help

  1. #1
    Junior Member
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    algebra inequality help

    Hi, could you please help me solve this question? Thank you!



    If x and y are positive, which of the following must be greater than 1/sqrt(x+y) ?

    1) (sqrt(x+y))/(2x)

    2) (sqrt(x)+sqrt(y))/(x+y)

    2) (sqrt(x)-sqrt(y))/(x+y)
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  2. #2
    MHF Contributor red_dog's Avatar
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    1) \displaystyle\frac{\sqrt{x+y}}{2x}>\frac{1}{\sqrt{  x+y}}\Leftrightarrow x+y>2x\Leftrightarrow y>x
    So, the inequality is true iff y>x and not for all positive x and y.

    2) \displaystyle\frac{\sqrt{x}+\sqrt{y}}{x+y}>\frac{1  }{\sqrt{x+y}}\Leftrightarrow\frac{\sqrt{x}+\sqrt{y  }}{x+y}>\frac{\sqrt{x+y}}{x+y}\Leftrightarrow\sqrt  {x}+\sqrt{y}>\sqrt{x+y}\Leftrightarrow 2\sqrt{xy}>0
    which is true for all positive x, y.

    3) \displaystyle\frac{\sqrt{x}-\sqrt{y}}{x+y}>\frac{1}{\sqrt{x+y}}
    Is not true if x<y, because LHM is negative and RHM is positive.
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