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From Spivak's calculus 3rd edition:

The result in Problem 1-7 has an important generalization: If $a_1,...,a_n \geq 0$, then the "arithmetic mean" $$A_n = \frac{a_1+...+a_n}{n}$$ and "geometric mean" $$G_n = \sqrt[n]{a_1...a_n}$$ satisy $$G_n \leq A_n$$(a) Suppose that $a_1 < A_n$ Then some $a_i$ satisfies $a_i>A_n$; for convenience, say $a_2>A_n$. Let $\bar{a_1} = A_n$ and let $\bar{a_2} = a_1 + a_2 - \bar{a_1}$. Show that $$\bar{a_1}\bar{a_2} \geq a_1a_2$$ Why does repeating this process enough times eventually prove that $G_n \leq A_n$?

I've got the first part but I'm struggling to understand the "repeating process" aspect of the proof. From what i understand it comes from each iteration of the process increasing $G_n$ but keeping $A_n$ constant but i don't get how exactly the process is repeated.

For example, Given a set of values $\{a_1, a_2\}$,
if $$a_1 < A_2 < a_2$$ $$\bar{a_1}=A_2\\ \bar{a_2}=a_1+a_2-\bar{a_1}$$

then a new set $\{\bar{a_1}, \bar{a_2}\}$ can be created where $\bar{A_2} = A_2$ and $\bar{G_n} \geq G_n$
How would you repeat the process for $\{\bar{a_1}, \bar{a_2}\}$?

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The set that you're working with always has $n$ elements, but at each step you change only two at a time.

If all of the elements are equal to the arithmetic average $A_n$, then we see $$G_n = \sqrt[n]{(A_n)^n} = A_n.\tag{1}\label{1}$$

On the other hand, if not all of the elements are equal to $A_n$ then there must be at least one that is less than $A_n$, and likewise, at least one that is greater than $A_n$. (Can you see why?)

In this case, by assumption, we have $a_1$ and $a_2$ with $$a_1 < A_n < a_2.$$

For the first step, we replace $a_1$ and $a_2$ with $\bar{a_1}$ and $\bar{a_2}$, where $$\bar{a_1} = A_n,$$ and $$\bar{a_2}=a_1+a_2-\bar{a_1}.$$

Some arithmetic will show you that the set formed with these new elements $\{\bar{a_1}, \bar{a_2}, a_3,\dots, a_n\}$, has the same arithmetic mean as the original set $\{a_1, a_2,\dots, a_n\}$ and a geometric mean that is greater than that of the original set. $$A_n' = A_n,$$ $$G_n' > G_n.$$

If all of the elements of this new set are equal to $A_n'$ then combining the above and \eqref{1}, we have $$G_n < G_n' = A_n' = A_n.$$

On the other hand, if there is at least one element of the new set that is less than $A_n$, we can repeat the process, this time calling the two elements $a_2$ and $a_3$, with

$$a_2 < A_n < a_3.$$

Note that $\bar{a_2}$ from the first step might be either one of these numbers. $\bar{a_1}$ on the other hand, cannot be either. (Remember, we set $\bar{a_1} = A_n$. We are done messing with $\bar{a_1}$.)

So, if this second step is needed, we replace the above numbers $a_2$ and $a_3$ with $\bar{a_2}$ and $\bar{a_3}$, where $$\bar{a_2} = A_n,$$ and $$\bar{a_3}=a_2+a_3-\bar{a_2}.$$

The set formed with these new elements $\{\bar{a_1}, \bar{a_2}, \bar{a_3}, a_4, \dots, a_n\}$ has the same arithmetic mean as the previous set, which has the same arithmetic mean as the original set. Our new set has a greater geometric mean than the previous set, which has a greater geometric mean than that of the original set.

We continue in this way until every element equals $A_n$. How do we know we will eventually reach this state? Each round in the process takes two elements that are not equal to $A_n$ and replaces them with two elements, at least one of which equals $A_n$. There is at least one less "bad" element not equal to $A_n$ after each round.

Eventually, all of the elements will equal $A_n$ and equation \eqref{1} will apply.

A helpful exercise: assume the set has only two "bad" elements $a_1$ and $a_2$ (with all other $a_i = A_n$) and proceed with the prescribed substitutions: replace $a_1$ and $a_2$ with $\bar{a_1}$ and $\bar{a_2}$, where $$\bar{a_1} = A_n,$$ and $$\bar{a_2}=a_1+a_2-\bar{a_1}.$$

In this case, how does $\bar{a_2}$ compare with $A_n$? (Hint: remember that the sum of the two elements and hence the arithmetic mean of the set remains unchanged.)

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