AM - GM inequality with Bernoulli's inequality

Question

I am stuck on this proof: The AM-GM Inequality is Equivalent to the Bernoulli Inequality.

I get the first part with the Bernoulli Inequality, but how did he get from $\frac{A_n}{A_{n-1}}\ge \frac{x_n}{A_{n-1}}$ to $A_n^n\ge x_{n}A_{n-1}^{n-1}$
Can someone explain to me what I am missing there?

Thanks in advance!

Answer 1

The initial inequality is not $\frac{A_n}{A_{n-1}} \ge \frac{x_n}{A_{n-1}}$ but $$\left(\frac{A_n}{A_{n-1}}\right)^n \ge \frac{x_n}{A_{n-1}}.$$ Multiplying both sides by $A_{n-1}^n$, we get $$ \left(\frac{A_n}{A_{n-1}}\right)^n A_{n-1}^n \ge \frac{x_n}{A_{n-1}} \cdot A_{n-1}^n \implies A_n^n \ge x_n \cdot A_{n-1}^{n-1}. $$