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'?
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.
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.
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."
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