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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!

RobPratt
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dragontec12
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  • For other proofs that Bernoulli implies AM-GM see https://math.stackexchange.com/questions/4198527/chrystals-proof-of-the-arithmetic-mean-geometric-mean-inequality and https://math.stackexchange.com/questions/3811003/proof-by-induction-of-am-gm-inequality – marty cohen Nov 18 '22 at 04:03

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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 $$ \left(\frac{A_n}{A_{n-1}}\right)^n A_{n-1}^n \ge \frac{x_n}{A_{n-1}} \cdot A_{n-1}^n \implies A_n^n \ge x_n \cdot A_{n-1}^{n-1}. $$

Misha Lavrov
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