# Geometric Mean is Never Less than Harmonic Mean

## Theorem

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

Let $G_n$ be the geometric 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 $G_n \ge H_n$.

## Proof

Let ${G_n}'$ denotes the geometric mean of the reciprocals of $x_1, x_2, \ldots, x_n$.

By definition of harmonic mean, we have that:

$\dfrac 1 {H_n} = \displaystyle \sum_{k \mathop = 0}^n \dfrac 1 {x_n}$

That is, $\dfrac 1 {H_n}$ is the arithmetic mean of the reciprocals of $x_1, x_2, \ldots, x_n$.

Then:

 $\displaystyle \dfrac 1 {H_n}$ $\ge$ $\displaystyle {G_n}'$ Cauchy's Mean Theorem $\displaystyle$ $=$ $\displaystyle \dfrac 1 {G_n}$ Geometric Mean of Reciprocals is Reciprocal of Geometric Mean $\displaystyle \leadsto \ \$ $\displaystyle H_n$ $\le$ $\displaystyle G_n$ Reciprocal Function is Strictly Decreasing

$\blacksquare$