I'm trying to solve the following question which leads to proving the AM-GM inequality.
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"
$$G_n = \sqrt[n]{a_1…a_n}$$
satisfy
$$G_n \leq A_n$$
(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 $G_n \leq A_n$? (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 $G_n \leq A_n$?
I have looked up in the Wikipedia about proofs for AM-GM inequality and it seems like this question wants me to prove it using the method of what's called "successive replacement of elements" according to the Wikipedia (https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means).
I'm having trouble understanding the point of setting $\bar{a_2} = a_1 + a_2 – \bar{a_1}$.
Doing this is suppose to leave the arithmetic mean unchanged but geometric mean increased but I don't see how this works. Why are we suddenly saying $A_n = \bar{a_1}$ and setting $\bar{a_2} = a_1 + a_2 – \bar{a_1}$? Also I don't think I understand the exact meaning of the symbols here. What does saying $a_1 < A_n < a_2$ imply? Is $A_n$ the arithmetic mean of elements that are in-between $a_1$ and $a_2$ but not including $a_1, a_2$?
Please ask me if anything is unclear. Thank you in advance.
Best Answer
By replacing the sequence $a_1,...,a_n$ by $\overline a_1, \overline a_2, ..., a_n$ (with $\overline a_1 = A_n$ and $\overline a_2 = a_1 + a_2 - A_n$), the arithmetic mean $A_n$ is unchanged because $$A_n = \frac{\overline a_1+ \overline a_2+ ...+ a_n} n = \frac{a_1+ a_2+ ...+ a_n} n $$ but the geometric mean increases: $$ \big(\overline a_1 \overline a_2 ... a_n \big)^{1/n} \geq \big( a_1 a_2 ... a_n \big)^{1/n} $$ since $\overline a_1 \overline a_2 \geq a_1 a_2$. If you repeat the process until no element of the sequence is $< A_n$, you end up with a constant sequence for which arithmetic and geometric means are equal.