Use induction to prove the AM-GM inequality
My attempt:
For n=1, we have $\displaystyle a_1=a_1$, which is trivially true.
Let the statement be true for some n.
$\displaystyle \frac{a_1+a_2+...+a_n}{n}\geq(a_1a_2...a_n)^{1/n}$
$\displaystyle \Leftrightarrow a_1+a_2+...+a_n\geq n(a_1a_2...a_n)^{1/n}$
We need to prove that $\displaystyle a_1+a_2+...+a_{n+1}\geq(n+1)(a_1a_2...a_{n+1})^{1/(n+1)}$
We know that $\displaystyle a_1+a_2+...+a_{n+1}\geq n(a_1a_2...a_n)^{1/n}+a_{n+1}$
How do I show that $\displaystyle n(a_1a_2...a_n)^{1/n}+a_{n+1}\geq (n+1)(a_1a_2...a_{n+1})^{1/(n+1)}$?
Re: Use induction to prove the AM-GM inequality
