Harmonic Mean of two Positive Real Numbers is Between them

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.

$\blacksquare$