GM-HM Inequality/Proof 1

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:

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

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

$\blacksquare$