How do the Bernoulli's inequality gives the AM-GM?

Question

I heard that one can use Bernoulli's inequality to prove the arithmetic-geometric inequality. However, I was unable to find the details of the proof. Does anyone knows how to prove it? I guess it is in http://link.springer.com/article/10.1007/s00283-011-9266-8 but I have no access to the article.

Answer 1

Check http://www.researchgate.net/publication/257393619_The_AM-GM_Inequality_is_Equivalent_to_the_Bernoulli_Inequality

The relevant part is, defining $A_k$ to be the arithmetic mean of the first $k$ terms of $x_1, x_2, \dots, x_n$, $$\left(1+\frac{A_n}{A_{n-1}}-1 \right)^n \ge 1+n\left(\frac{A_n}{A_{n-1}}-1\right) = \frac{nA_n - (n-1)A_{n-1}}{A_{n-1}} = \frac{x_n}{A_{n-1}}$$

Using $A_n^n \ge x_n A_{n-1}^{n-1}$ iteratively, one gets $$A_n^n \ge x_n x_{n-1} \dots x_1$$ which is AM-GM.