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Inequality of arithmetic and geometric means

A.M. = \frac{ a_1 + a_2 + a_3 + \ldots + a_n}{n} \: and \: G.M. = \sqrt[n]{ a_1 \cdot a_2 \cdot a_3 \cdots a_n } . Let's \: proof \: A.M. \ge G.M. \: a_i \: is \: non-negative \: Real \: number.

First of all we need to proof that (a_1 + a_2 )/2 \ge \sqrt{ a_1 \cdot a_2 } It is easy, you just have to know ( \sqrt{a_1} + \sqrt{a_2})^2 \ge 0

Let’s accept that this inequality is true for n=k and proof that it is true for n=2k. ( k is positive integer )

( a_1 + a_2 + \ldots + a_{2k} )/{2k} = ( \frac{ a_1 + a_2 }{2} + \frac{ a_3 + a_4 }{2} + \cdots + \frac{ a_{2k-1} + a_{2k} }{2} ) /k \ge

\ge \sqrt[k]{ \frac{ a_1 + a_2 }{2} \cdot \frac{ a_3 + a_4 }{2} \cdots \frac{ a_{2k-1} + a_{2k} }{2} } \ge \sqrt[2k]{ a_1 \cdot a_2 \cdots a_{2k} }

For n=2 inequality is true, then n = 2^m inequality is true, too. ( m \in {\cal N } )

So, let n + p = 2^m \: ( p \in {\cal N } ) \: and \: a_{n+1} = a_{n+2} = \cdots = a_{n+p} = \big( a_1 +a_2 + \cdots + a_n \big) /n .

\bigg( \sum \limits_{i=1}^{n+p} a_i \bigg) /{ (n+p) } \ge \sqrt[n+p]{ \prod \limits_{i=1}^n a_i \cdot a_{n+1}^p }

\frac{ a_1 + \cdots + a_n + a_{n+1} + \dots + a_{n+p} }{n+p} = \frac{ a_1 + \cdots + a_n + p \cdot a_{n+1} }{n+p} = \bigg( a_1 + a_2 + \cdots + a_n \bigg) /n

\sqrt[n+p]{ ( \prod \limits_{i=1}^n a_i ) \cdot a_{n+1}^p } = \bigg( \prod \limits_{i=1}^n a_i \bigg)^{1/{n+p}} \cdot \bigg( \frac{ \sum \limits_{i=1}^n a_i }{n} \bigg)^{ p/(n+p) } \Rightarrow

( a_1 + a_2 + \cdots + a_n )/n \ge \bigg( \prod \limits_{i=1}^n a_i \bigg)^{1/{n+p}} \cdot \bigg( \frac{ \sum \limits_{i=1}^n a_i }{n} \bigg)^{ p/(n+p) }

( a_1 + a_2 \cdots + a_n )/n \ge \sqrt[n]{ a_1 \cdot a_2 \cdots a_n }

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