How do you prove that an arithmetic average is greater than or equal to a geometric average? This the same as showing that zⁿ > = nz for all n natural numbers, and z greater than 1.

Thanks,

Brad

I think we have done that before... Anyway, there are a few proofs. Here is how one works:

Let P(n) be the statement
(a₁ +...+a_n )/n > = (a₁ ...a_n )^1/n

When n=1, the case is immediate.
When n=2, recall (sqrt(a₁ )-sqrt(a₂ ))² > = 0 and simply rearrange to give the required inequality.
When n=2^k , k> 1, we can show, by repeated use of the case n=2, that the inequality holds. e.g. let b_r =(a_2r-1 +a_2r )/2 and the induction step is immediate.

This leaves to prove that if P(n) is true then P(n-1) is also true. Substitute a_n =(a₁ +...+a_n-1 )/(n-1) into P(n) and after a bit of rearranging, the result follows.

Sorry about the briefness but do write back if you are stuck.

Kerwin

By the way, this is not the same as zⁿ > = nz for natural n and z greater than 1. In fact that is false in general. To be true we require z^n-1 > = n for all z. But as z-> 1 the LHS-> 1 disproving the inequality for n> 1.

Another way of proving AM-GM is to simply consider:

y = [(a₁ + ... + a_n )/n]/[a₁ ...a_n ]^(1/n)

and then differentiate y with respect to a_n leaving all else constant, and set this to 0 to find minimum. The result you get immdiately shows the fraction is minimised when a₁ = a₂ = ... = a_n . Then equality occurs.