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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- Jul 5th 2012, 10:25 AM #1

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- Jul 5th 2012, 10:35 AM #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.

- Jul 5th 2012, 11:00 AM #3

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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.

- Jul 5th 2012, 05:29 PM #4

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- Jul 6th 2012, 09:45 AM #5

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- Jul 6th 2012, 09:55 AM #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."

- Jul 8th 2012, 05:58 AM #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