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Thread: 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,

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

    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

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

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

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

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

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

    Challenge: Prove that for all nonnegative x and y that

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

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