Vtyjkyuk

I am Anay. I was reading the second chapter "Numbers of Various Sorts" of Spivak Calculus 4th Edition when I came accross this, in Problem 22:

The result in Problem 1-7 has an important generalization: If $a_1, ..., a_n geq 0$, then the "arithmetic mean"
$$A_n = fraca_1+...+a_nn$$
and "geometric mean"
$$G_n = sqrt[n]a_1...a_n$$
satisy
(a) Suppose that $a_1 < A_n$. Then some $a_i$ satisfies $a_i > A_n$; for convenience, say $a_2 > A_n$. Let $bara_1 = A_n$ and let $bara_2 = a_1 + a_2 - bara_1$. Show that $$bara_1bara_2 geq a_1a_2.$$
Why does repeating this process enough times eventually prove that $G_n leq A_n$? (This is another place where it is a good exercise to provide a formal proof by induction, as well as an informal reason.) When does equality hold in the formula $G_n leq A_n$?

This is exactly what was written in the book. Now I showed $bara_1bara_2 geq a_1a_2$, that was easy. but I don't see how "repeating" this process enough times proves $G_n leq A_n$. What are we supposed to repeat exactly? What should we take as $bara_3, bara_4...$? I tried this in a variety of ways but I am not able to construct a proof. I tried searching for AM-GM proofs online to see if such a proof is listed and I couldn't find this one. So, I am asking this question here. How do I complete this AM-GM proof?

Thanks in advance.

What you have shown is the following:

If we replace $(a_1,...,a_n)$ by $(bara_1,bara_2,a_3,...,a_n)$ then the arithmetic mean of both $n$-tuples stays the same, while the geometric mean of the latter hasn't decreased relative to the former. Furthermore, in the $n$-tuple $(bara_1,bara_2,a_3,...,a_n)$ there are strictly more numbers that are equal to $A_n$ than in the $n$-tuple $(a_1,...,a_n)$, because we supposed $a_1<A_n<a_2$. So as long as there is an $i$ such that $a_i<A_n$, we can use the same argument as above to strictly increase the number of numbers in the $n$-tuple equal to $A_n$, whitout decrasing the corresponding geometric mean. This is what is meant by repeating this procedure.

Can you see what all this leads to? If no, say so, and I will further elaborate.

Explanation (for @Honda):

The operation of replacing $(a_1,...,a_n)$ by $(bara_1,bara_2,a_3,...,a_n)$ doesn't affect the arithmetic mean. Furthermore, as $bara_1bara_2geq a_1a_2$, the geometric mean increases. As pointed out above, we can repeat this operation until all of the $a_i$ are equal to $A_n$. At this point, the geometric mean also equals $A_n$. As it only increased while repeating the procedure, the initial geometric mean $G_n$ must be smaller than the final one, that is $A_n$. We thus proved $G_nleq A_n$.