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Math Help - More with AM and GM

  1. #1
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    More with AM and GM

    IF a,b >0
    prove: AM - GM >= GM - HM
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  2. #2
    Super Member Rebesques's Avatar
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    If I get it right, u need to show that for all a,b>0,

    \frac{a+b}{2}-\sqrt{ab}\geq\sqrt{ab}-\frac{2}{\frac{1}{a}+\frac{1}{b}}.

    Both sides are positive; Begin calculating until you get to a valid inequality.
    (remember AM>GM>HM.)
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  3. #3
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    Cool

    \frac{a+b}{2}-\sqrt{ab}\ge \sqrt{ab}-\frac{2}{\frac{1}{a}+\frac{1}{b}}

    \frac{a+b}{2}-\sqrt{ab}\ge \sqrt{ab}-\frac{2ab}{a+b}

    \frac{a+b}{2}+\frac{2ab}{a+b}\ge 2\sqrt{ab}

    from here we just have to apply AM-GM. Note that \left(\frac{a+b}{2}\right)\left(\frac{2ab}{a+b}\ri  ght)=ab.

    if you don't know AM-GM, basically it just states: for any x,y\ge 0 we are assured that x+y\ge 2\sqrt{xy}, with equality iff (if and only if) x=y.

    Challenge: Prove that for all nonnegative x and y that

    \sqrt{\frac{x^2+y^2}{2}}-\frac{x+y}{2} \ge \sqrt{xy}-\frac{2}{\frac{1}{x}+\frac{1}{y}}

    try not to use the fact that \sqrt{\frac{x^2+y^2}{2}}\ge\frac{x+y}{2}
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