Proof of AM-GM Inequality: Spivak's Calculus Chapter 2 Problem 22 Part A

[erh5060]

Homework Statement

From Spivak's Calculus, Chapter 2 Problem 22 Part A:

Here, $$A_{n}$$ and $$G_{n}$$ stand for the arithmetic and geometric means respectively and $$a_{i}\geq 0$$ for $$i=1,\cdots,n$$.

Suppose that $$a_{1} < A_{n}$$. Then some $$a_{i}$$ satisfies $$a_{i} > A_{n}$$; for convenience, say $$a_{2} > A_{n}$$. Let $$a^{*}_{1} = A_{n}$$, and let $$a^{*}_{2} = a_{1} + a_{2} - a^{*}_{1}$$. Show that

$$a^{*}_{1} a^{*}_{2} \geq a_{1} a_{2}$$

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 $$G_{n} \leq A_{n}$$?

Homework Equations

The Attempt at a Solution

The first part of the proof was easy:

$$A^{2}_{n} - (a_{1} + a_{2}) A_{n} + a_{1} a_{2} = (A_{n} - a_{1})(A_{n} - a_{2}) < 0$$ since it was given that $$a_{1} < A_{n} < a_{2}$$. Rearranging this leads to $$a_{1} a_{2} < a^{*}_1 a^{*}_2$$, which is what the problem asked for.

It is also clear that equality holds when $$a_{1} = a_{2} = \cdots = a_{n} = A_{n}$$. The part that is baffling me is filling in the rest of the logic of this proof to show that $$G_{n} \leq A_{n}$$. I'm not exactly sure what process I'm suppose to "repeat" as stated in the problem, or how I'm supposed to turn it into a formal proof by induction. None of $$a^{*}_{i}$$ for $$i \geq 3$$ were defined anywhere in the problem. Any help would be appreciated.

 

[snipez90]

Hmm, well you basically already proved that the GM of the a_i's is less than the GM of the $a^*_i$'s. But note that the AM of the a_i's is equal to the AM of the $a^*_i$'s, so perhaps try to exploit this fact.