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Math Help - Which is Greater, inequality?

  1. #1
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    Which is Greater, inequality?

    (x_1 + x_2 + ... + x_n)/n *** (x_1*x_2*...*x_n)^(1/n)

    Fill in *** with >, <, >=, <= and please explain how you came to that conclusion.

    Thanks
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  2. #2
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    Quote Originally Posted by Ideasman View Post
    (x_1 + x_2 + ... + x_n)/n *** (x_1*x_2*...*x_n)^(1/n)

    Fill in *** with >, <, >=, <= and please explain how you came to that conclusion.

    Thanks
    If x_i>=0 then the first one is greater.

    That is the AM-GM inequality.
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  3. #3
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    the ph has already explained the proof for this in a previous thread. Ill re explain it for you.

    Arithmetic mean:  AM(a_1,....,a_n) = \frac{a_1 + .... + a_n}{n}

    Geometric mean: [tex] GM(a_1,.....a_n) = \sqrt[n]{a_1,....,a_n}

    Proof by induction:

    without loss of generality we may resclae a_i so that a_1.....a_n = 1 . If all a_i= 1 the proof is trivial.

    Thus assume atleast one a_i > 1 and one a_i < 1. We assume a_1 > 1, a_2 < 1. By inductive assumption we have

     \frac{a_1a_2 + a_3 + ......+ a_n}{n-1} \geq \sqrt[n-1]{(a_1 a_2) a3.....a_n = 1

    thus we have

    a_1a_2+a_3+....a_n \geq n-1

    we need to show

    a_1 + a_2+a_3+....a_n \geq n

    this would follow if a_1 + a_2 - (a_1a_2 + 1) \geq 0 BUT


    a_1 + a_2 - (a_1a_2 + 1) = (a_1 - 1)(1-a_2) \geq 1,

    prooving what you have writen
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  4. #4
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    hopefully that makes sense even tho latex is bust. if not ill pdf it for you
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  5. #5
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    Quote Originally Posted by chogo View Post
    hopefully that makes sense even tho latex is bust. if not ill pdf it for you
    That no makes no sense to me.

    It seems you made this inequality look too easy.
    The standard method is by max/minimizing the multivarible function. Or by Lagrange multipliers.

    However! The cleanest proof uses the best inequality (the Cauchy-Swartzh Inequality).
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  6. #6
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    i thought that was the standard induction proof. Yeah i liked the cauchy proof (the one u did in the previous thread for n=2). Though i would try this tho
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  7. #7
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    If you have the proof in PDF (since LateX is down) I'd appreciate it if you could post it. I just need a simple explanation.

    Thanks.
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  8. #8
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    Quote Originally Posted by ThePerfectHacker View Post
    If x_i>=0 then the first one is greater.

    That is the AM-GM inequality.
    x_i?
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  9. #9
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    If you have access to a reasonable mathematics library, you may find AN INTRODUCTION TO INEQUALITIES by Beckenbach & Bellman very useful. Thy have a rather complete discussion of this topic in that book.
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  10. #10
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    http://www.math.princeton.edu/mathla...eanGeoMean.pdf

    thats the one. PH what do you think? is that wrong? have you got a link for the maximisation proof?
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