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.