Maximum is Greater than or Equal to Geometric Mean

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

Let $p \in \R$ be a real number.

Let $\map G {x_1, x_2, \ldots, x_n}$ denote the geometric mean of $x_1, x_2, \ldots, x_n$.

Then:
 * $\max \set {x_1, x_2, \ldots, x_n} \ge \map G {x_1, x_2, \ldots, x_n}$

Equality holds :
 * $x_1 = x_2 = \cdots x_n$

or $x_k = 0$ for some $k \in \set {1, 2, \ldots, n}$.

Proof
From Limit of Hölder Mean as Exponent tends to Zero is Geometric Mean:


 * $\map G {x_1, x_2, \ldots, x_n} = \ds \lim_{p \mathop \to 0} \map {M_p} {x_1, x_2, \ldots, x_n}$

where $\map {M_p} {x_1, x_2, \ldots, x_n}$ denotes the Hölder mean with exponent $p$ of $x_1, x_2, \ldots, x_n$.

Then from Maximum is Greater than or Equal to Hölder Mean:

$\max \set {x_1, x_2, \ldots, x_n} \ge \map {M_p} {x_1, x_2, \ldots, x_n}$

where equality holds :
 * $x_1 = x_2 = \cdots x_n$

or:
 * $p < 0$ and $x_k = 0$ for some $k \in \set {1, 2, \ldots, n}$.

In this context, if such an $x_k = 0$ then:
 * $\ds \prod_{k \mathop = 1}^n x_k = 0$

and so by definition: $\map G {x_1, x_2, \ldots, x_n} = 0$

Thus we have:
 * $\max \set {x_1, x_2, \ldots, x_n} \ge \map G {x_1, x_2, \ldots, x_n}$