Harmonic Mean of two Real Numbers is Between them

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Let $a, b \in \R_{\ne 0}$ be non-zero real numbers such that $a < b$.

Let $\map H {a, b}$ denote the narmonic mean of $a$ and $b$.


$a < \map H {a, b} < b$


By definition of harmonic mean:

$\dfrac 1 {\map H {a, b} } := \dfrac 1 2 \paren {\dfrac 1 a + \dfrac 1 b}$


\(\ds a\) \(<\) \(\ds b\) by assumption
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 b\) \(<\) \(\ds \dfrac 1 a\) Reciprocal Function is Strictly Decreasing

But $\dfrac 1 {\map H {a, b} }$ is the arithmetic mean of $\dfrac 1 b$ and $\dfrac 1 a$.

Hence from Arithmetic Mean of two Real Numbers is Between them:

$\dfrac 1 b < \dfrac 1 {\map H {a, b} } < \dfrac 1 a$

So by Reciprocal Function is Strictly Decreasing:

$b > \map H {a, b} > a$

Hence the result.