Chapter 8.1, Problem 53HP

To determine

Toexplain:

Condition when arithmetic and geometric means will be equal.

Answer to Problem 53HP

Condition when arithmetic and geometric means will be equal is $a=b$ that is $\frac{a+b}{2}\ge \sqrt{ab}$ .

Explanation of Solution

Given information:

Arithmatic and geometric means of two numbers.

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

The geometric mean of $a$ and $b$ is $\sqrt{ab}$ .

Suppose, that both $a$ and $b$ are positive numbers. Then, both $\frac{a+b}{2}$ and $\sqrt{ab}$ are no less than the smaller one and no greater than the greater one.

So, assume without loss of generously that $a\le b$ . Then, $a\le \sqrt{ab}\le \frac{a+b}{2}\le b$

The equality occurs if and only if $a=b$

Analyze the difference between $\frac{a+b}{2}$ and $\sqrt{ab}$

  $\begin{array}{l}\frac{a+b}{2}-\sqrt{ab}=\frac{1}{2}\left(a+b-2\sqrt{ab}\right)\\ =\frac{1}{2}{\left(\sqrt{a}-\sqrt{b}\right)}^2\\ \ge 0\end{array}$

This means that $\frac{a+b}{2}\ge \sqrt{ab}$ and that the equality occurs if and only if $a=b$ .