1. 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'?

2. 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.

3. 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.

4. 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.

5. 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.

6. 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."

7. 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