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:


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

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

Then: