If a₁ = a₂ = ... = a_k+1 then A_k+1 = a₁,G_k+1 = a₁, and so  It remains to investigate the case in which the a_i are not all equal. Without loss of generality, we may assume that the a_i are numbered so that

The fact that the a's are not all equal implies that a₁ < a_k+1. It now follows from Problem 18 of Section 10.1 that a_k+1 > A_k. Since

we have kA_k = a₁ + a₂ + ... + a_k, and hence

Let (a_k+1 - A_k)/(k + 1) = p. We have seen above that a_k+1 > A_k; this implies that p > 0. Now A_k+1 = A_k + p. We raise both sides of this equality to the (k + 1)th power, obtaining

Since p > 0 and A_k > 0, all the terms in the binomial expansion on the right side are positive. There are k + 2 terms in this expansion; hence there are at least 4 terms. Now (A_k+1)^k+1 is greater than the sum of the first two terms:

Since (k + 1)p = a_k+1 - A_k, this becomes

Having assumed above that we now have

From Problems 7 and 8, Section 10.1, it follows that A_k+1 > G_k+1, and the theorem is proved. We have actually proved more than is stated in the theorem; we have shown that A_n > G_n unless all the a_i are equal.
 
 

Monday, June 22, 1998