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Math Help - A.M. and G.M. help

  1. #1
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    A.M. and G.M. help

    How can i prove, 'Arithmetic Mean and Geometric Mean of n positive reals are equal' implies that 'all the n positive reals are the same'?
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  2. #2
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    Re: A.M. and G.M. help

    this question is interesting. the direction is arithmetic mean is always no less then geometry mean. they equal if all positive nums are equal. this is a well-known inequality, called cauchy inequality if im not wrong.

    my hint is as follow (using mi)
    for 2 numbers: (a+b)/2 >= sqrt(a+b)
    assume that (a1+a2+....+an)/n >= sqrt[n](a1xa2x....xan)
    prove that the inequality still hold for n+1

    during the proof, we can easily see that "=" happens when all num are equal.
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    Re: A.M. and G.M. help

    I saw there is this statement in some math books,"Arithmetic Mean and Geometric Mean of n positive reals are equal' iff 'all the n positive reals are the same" . I have already worked out the proof from right to left. And want to know the proof from left to right.
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  4. #4
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    Re: A.M. and G.M. help

    what u mean by left to right? in fact there are lots of ways to solve actually. we can put up here to discuss and broaden our knowledge. please give me sometime i will post solution tonite.
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  5. #5
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    Re: A.M. and G.M. help

    Don't u know what 'iff' means? If u know, u can understand the kind of question i'm asking.
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  6. #6
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    Re: A.M. and G.M. help

    AoPS has some good proofs for AM-GM: Arithmetic Mean-Geometric Mean Inequality - AoPSWiki. The equality cases for when AM = GM should follow.

    Just in case you don't know, "iff" means "if and only if."
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  7. #7
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    Re: A.M. and G.M. help

    here is the link for the proof using mi, which take care of the iff thing altogether.

    Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia
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