Harmonic Mean of two Positive Real Numbers is Between them
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Theorem
Let $a, b \in \R_{\gt 0}$ be (strictly) positive real numbers such that $a < b$.
Let $\map H {a, b}$ denote the harmonic mean of $a$ and $b$.
Then:
- $a < \map H {a, b} < b$
Proof
By definition of harmonic mean:
- $\dfrac 1 {\map H {a, b} } := \dfrac 1 2 \paren {\dfrac 1 a + \dfrac 1 b}$
Thus:
\(\ds a\) | \(<\) | \(\ds b\) | by hypothesis | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 b\) | \(<\) | \(\ds \dfrac 1 a\) | 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.
$\blacksquare$