# AM-GM Proof in Spivak Calculus

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" $$A_n = \frac{a_1+...+a_n}{n}$$ 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 $\bar{a_1} = A_n$ 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 $G_n \leq A_n$? (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 $\bar{a_1}\bar{a_2} \geq a_1a_2$, 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 $\bar{a_3}, \bar{a_4}...$? 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?

• I think that $\overline a_3$ will be $a_1 + a_2 + a_3 - \overline a_2$ and that we have to show that: $\overline a_1 \overline a_2 \overline a_3 \ge a_1 a_2 a_3$. Oct 20, 2017 at 13:18

What you have shown is the following:

If we replace $(a_1,...,a_n)$ by $(\bar{a}_1,\bar{a}_2,a_3,...,a_n)$ 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 $(\bar{a}_1,\bar{a}_2,a_3,...,a_n)$ 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 $(\bar{a}_1,\bar{a}_2,a_3,...,a_n)$ doesn't affect the arithmetic mean. Furthermore, as $\bar{a}_1\bar{a}_2\geq a_1a_2$, 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 $G_n\leq A_n$.

• @AnayKarnik You're welcome :) Oct 21, 2017 at 20:55
• @RedundantAunt I can't see where this leads to, may you please elaborate Aug 26, 2018 at 14:25
• @Honda I added an explanation; let me know if something is still unclear. Aug 29, 2018 at 8:11
• @RedundantAunt Thanks! Sep 5, 2018 at 3:24
• Hi @RedundantAunt. So $\bar{a_{i}} = a_{i}$ for $3 \leq i \leq n$? Feb 12, 2021 at 0:03