Chapter A.9, Problem 126AYU

To determine

To find: to show that harmonic mean of $a$ and $b$ equals the geometric mean squared divided by the arithmetic mean.

Answer to Problem 126AYU

The verification is shown below.

Explanation of Solution

Given information:

It is given that $a<b$ and $\frac{a+b}{2}$ is arithmetic mean of $a$ and $b$ . It is given that $0<a<b$ and $\sqrt{ab}$ is geometric mean of $a$ and $b$ . The harmonic mean is $\frac{1}{h}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)$ and $0<a<b$ .

Calculation:

The arithmetic mean of $a$ and $b$ is $\frac{a+b}{2}$ and the geometric mean is $\sqrt{ab}$ and The harmonic mean is $\frac{1}{h}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)$

Simplify harmonic mean.

  $\begin{array}{l}\frac{1}{h}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\\ \frac{1}{h}=\frac{a+b}{2ab}\\ h=\frac{{\left(\sqrt{ab}\right)}^2}{\frac{a+b}{2}}\\ \text{harmonic}\text{mean}=\frac{{\left(\text{Geometric}\text{mean}\right)}^2}{\text{arithmetic}\text{mean}}\end{array}$

Assume that the geometric mean is less than the arithmetic mean

  $\begin{array}{c}\frac{a+b}{2}-\sqrt{ab}=\frac{a-2\sqrt{ab}+b}{2}\\ =\frac{{\left(\sqrt{a}-\sqrt{b}\right)}^2}{2}\end{array}$

The square of any number is greater than or equal to $0$ .

Thus,

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

Therefore, it is verified that harmonic mean of $a$ and $b$ equals the geometric mean squared divided by the arithmetic mean