# Spivak Chapter $2$ Problem $22$ Clarification

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 $$\{\bar{a_1}, \bar{a_2}\}$$?

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.)