Arithmetic mean is less than geometric mean (Spivak Calculus 3rd Chapter 2 Problem 22)

Let $A$ be the arithmetic mean (called $A_n$ in the question), and call a number $a_i$ unbalanced if $a_i \neq A$.

The arithmetic mean of the unbalanced elements is $A$ at all times during the execution of the algorithm. This makes every step of the process convert at least one unbalanced element to $A$, so that the number of unbalanced is eventually reduced to zero. Each step increases the product $a_1 a_2 \dots a_n$.

The inequality $G_n\leq A_n $ is a simple application of Jensen's inequality using the concavity of the $\log$ function.

Well, I found the corresponding question in my old Spivak, and I've got his "Supplement to Calculus" as well that contains all his answers. It was question 20 from chapter 2 in my old edition. Rather than convert it to LaTeX, I've taken the quicker route of posting images of the books! In my edition the question was as follows:

and the following 3 images are his answer

Hope that helps :-).

There's a wikipedia article on this which contains several proofs, so take your pick :-).