GM-HM Inequality

From ProofWiki
Jump to navigation Jump to search

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 1

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:

$\ds \dfrac 1 {H_n} := \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {x_k} }$

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


Then:

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

$\blacksquare$


Proof 2

For $p \in \R$, let $\map {M_p} {x_1, x_2, \ldots, x_n}$ denote the Hölder mean with exponent $p$ of $x_1, x_2, \ldots, x_n$.

By definition of Hölder Mean with $p = 0$:

$\map {M_0} {x_1, x_2, \ldots, x_n} = \map G {x_1, x_2, \ldots, x_n}$

From Hölder Mean for Exponent -1 is Harmonic Mean:

$\map {M_{-1} } {x_1, x_2, \ldots, x_n} = \map G {x_1, x_2, \ldots, x_n}$

The result follows from Inequality of Hölder Means:

$-1 < 0 \implies \map {M_{-1} } {x_1, x_2, \ldots, x_n} < \map {M_0} {x_1, x_2, \ldots, x_n}$

unless $x_k = 0$ for some $k \in \set {1, 2, \ldots, n}$ or $x_1 = x_2 = \cdots = x_n$, in which case:

$\map {M_{-1} } {x_1, x_2, \ldots, x_n} = \map {M_0} {x_1, x_2, \ldots, x_n}$

Hence the result.

$\blacksquare$


Also see


Sources