1
$\begingroup$

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?

Thanks in advance.

$\endgroup$
  • $\begingroup$ I think that $\overline a_3$ will be $a_1 + a_2 + a_3 - \overline a_2$ and that we have to show that: $\overline a_1 \overline a_2 \overline a_3 \ge a_1 a_2 a_3$. $\endgroup$ – Mauro ALLEGRANZA Oct 20 '17 at 13:18
2
$\begingroup$

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 $G_n\leq A_n$.

$\endgroup$
  • $\begingroup$ @AnayKarnik You're welcome :) $\endgroup$ – Redundant Aunt Oct 21 '17 at 20:55
  • $\begingroup$ @RedundantAunt I can't see where this leads to, may you please elaborate $\endgroup$ – Honda Aug 26 '18 at 14:25
  • $\begingroup$ @Honda I added an explanation; let me know if something is still unclear. $\endgroup$ – Redundant Aunt Aug 29 '18 at 8:11
  • $\begingroup$ @RedundantAunt Thanks! $\endgroup$ – Honda Sep 5 '18 at 3:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.