# Proof of AM-GM Inequality with lemmas

I need to prove the AM-GM Inequality using a few specific lemmas that I have already proven. I'm mostly just unsure what to do next and how to tie it all together at the end to finish the proof.

Here are the lemmas that I have already proven.

1. The theorem is true for $n=2$.
2. Suppose the statement is true for $n=k$. Then it is true for $n=2k$.

3. Suppose $m<n$ and $x_{1} ,...,x_{m}>0$. Then the geometric mean of n numbers is $g = \sqrt[m]{x_{1}*...*x_{m}}$.

4. Suppose $m<n$ and the theorem is true for any $n$ numbers. Then the theorem is true for any $m$ numbers.

Any help would be greatly appreciated!

## 1 Answer

Use induction with lemma 1 as a base case and lemma 2 to drive the induction step, proving AM-GM for any $n$ equal to a power of 2.

For any $m$ not a power of 2, AM-GM holds for some $n$ a power of 2 greater than $m$. By lemma 4, AM-GM holds for $m$ numbers.

• So I wouldn't necessarily need to use anymore principles or properties? The four lemmas I have already proven should be enough with your added explanation at the end? Sorry if I am misinterpreting! – ej313 Jun 15 '14 at 3:40
• @ej313: You've proven pretty much everything you need. – user2357112 Jun 15 '14 at 3:43