Chapter A.9, Problem 125AYU

To determine

To show: That $a < h < b$. The number h is called the harmonic mean of $a$ and $b$

Answer to Problem 125AYU

$a < h < b$

Explanation of Solution

Given:

$0 < a < b$, let h defined by $\frac{1}{h} = \frac{1}{2}\left(\frac{1}{a} + \frac{1}{b}\right)$.

Calculation:

$\begin{array}{l}\frac{1}{h} = \frac{1}{2}\left(\frac{1}{a} + \frac{1}{b}\right)\\ \frac{1}{h} = \frac{1}{2}\left(\frac{b + a}{ab}\right)\\ \frac{1}{h} = \frac{a + b}{2ab}\\ h = \frac{2ab}{a + b}\end{array}$

subtract a on both sides

$\begin{array}{l}h - a = \frac{2ab}{a + b} - a\\ h - a = \frac{2ab - a^2 + ab}{a + b}\\ h - a = \frac{ab - a^2}{a + b}\\ h - a = \frac{a\left(b - a\right)}{a + b} > 0\\ h - a > 0\end{array}$

$h > a$    -----(1)

similarly $b - h = b - \frac{2ab}{a + b}$

$\begin{array}{l}b - h = \frac{ba + b^2 - 2ab}{a + b}\\ b - h = \frac{b^2 - ab}{a + b}\end{array}$

$\begin{array}{l}b - h = \frac{b\left(b - a\right)}{a + b} > 0\\ b - h > 0\end{array}$

$b > h$   -----(2)

From (1) and (2)

$a < h < b$