Geometric Mean of two Positive Real Numbers is Between them

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:

\(\ds a\)

\(<\)

\(\ds b\)

\(\ds \leadsto \ \ \)

\(\ds a b\)

\(<\)

\(\ds b^2\)

\(\ds \leadsto \ \ \)

\(\ds \sqrt {a b}\)

\(<\)

\(\ds b\)

$\blacksquare$