Chapter A.9, Problem 124AYU

To determine

To find: to show that geometric mean of $a$ and $b$ is less than the arithmetic mean of $a$ and $b$ .

Answer to Problem 124AYU

The expression is represented as inequality as $21<x<30$ .

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 $b$ .

Calculation:

The arithmetic mean of $a$ and $b$ is $\frac{a+b}{2}$ and the geometric mean is $\sqrt{ab}$ .

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 the geometric mean is less than arithmetic mean