Geometric Mean of Reciprocals is Reciprocal of Geometric Mean

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Theorem

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

Let $G_n$ denote the geometric mean of $x_1, x_2, \ldots, x_n$.

Let ${G_n}'$ denote the geometric mean of their reciprocals $\dfrac 1 {x_1}, \dfrac 1 {x_2}, \ldots, \dfrac 1 {x_n}$.


Then:

${G_n}' = \dfrac 1 {G_n}$


Proof

\(\ds {G_n}'\) \(=\) \(\ds \paren {\prod_{1 \mathop = k}^n \dfrac 1 {x_k} }^{1/n}\) Definition of Geometric Mean
\(\ds \) \(=\) \(\ds \paren {\dfrac 1 {\prod_{1 \mathop = k}^n x_k} }^{1/n}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {\paren {\prod_{1 \mathop = k}^n x_k}^{1/n} }\)
\(\ds \) \(=\) \(\ds \dfrac 1 {G_n}\) Definition of Geometric Mean

$\blacksquare$