AM-HM Inequality/Proof 2

Theorem

Let $x_1, x_2, \ldots, x_n \in \R_{> 0}$ be strictly positive real numbers.

Let $A_n $ be the arithmetic mean of $x_1, x_2, \ldots, x_n$.

Let $H_n$ be the harmonic mean of $x_1, x_2, \ldots, x_n$.


Then $A_n \ge H_n$.


Proof

By Hölder Mean for Exponent 1 is Arithmetic Mean:

$A_n = \dfrac {x_1 + x_2 + \cdots + x_n} n = \map {M_1} {x_1, x_2, \ldots, x_n}$

By Hölder Mean for Exponent -1 is Harmonic Mean:

$H_n = \dfrac 1 {\dfrac 1 n \paren {\dfrac 1 {x_1} + \dfrac 1 {x_2} + \cdots + \dfrac 1 {x_n} } } = \map {M_{-1} } {x_1, x_2, \ldots, x_n}$

The result follows from Inequality of Hölder Means.

$\blacksquare$

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