Use induction to prove the AM-GM inequality

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August 11th 2011, 08:09 AM
alexmahone

Use induction to prove the AM-GM inequality

My attempt:

For n=1, we have $a_1=a_1$ , which is trivially true.
Let the statement be true for some n.

$\frac{a_1+a_2+...+a_n}{n}\geq(a_1a_2...a_n)^{1/n}$

$\Leftrightarrow a_1+a_2+...+a_n\geq n(a_1a_2...a_n)^{1/n}$

We need to prove that $a_1+a_2+...+a_{n+1}\geq(n+1)(a_1a_2...a_{n+1})^{1/(n+1)}$

We know that $a_1+a_2+...+a_{n+1}\geq n(a_1a_2...a_n)^{1/n}+a_{n+1}$

How do I show that $n(a_1a_2...a_n)^{1/n}+a_{n+1}\geq (n+1)(a_1a_2...a_{n+1})^{1/(n+1)}$ ?
August 11th 2011, 04:04 PM
Siron

Re: Use induction to prove the AM-GM inequality

You can find a proof using induction over here:
Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia

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