Geometric Mean is Never Less than Harmonic Mean

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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

$\blacksquare$


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