Geometric Mean of two Positive Real Numbers is Between them
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Theorem
Let $a, b \in \R$ be real numbers such that $0 < a < b$.
Let $\map G {a, b}$ denote the geometric mean of $a$ and $b$.
Then:
- $a < \map G {a, b} < b$
Proof
By definition of geometric mean:
- $\map G {a, b} := \sqrt {a b}$
where $\sqrt {a b}$ specifically denotes the positive square root of $a$ and $b$.
Thus:
\(\ds a\) | \(<\) | \(\ds b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^2\) | \(<\) | \(\ds a b\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a\) | \(<\) | \(\ds \sqrt {a b}\) |
and:
\(\ds a\) | \(<\) | \(\ds b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a b\) | \(<\) | \(\ds b^2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sqrt {a b}\) | \(<\) | \(\ds b\) |
$\blacksquare$