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 |
$\blacksquare$