Inequality

Problem:



Solution:

Let's start with the familiar arithmetic mean $ \ge $ geometric mean inequality.

$ \frac{a + b}{2} \ge \sqrt{ab} $.

Substitute $ a = x^2 $, $ b = \frac{4}{x^2} $, these must be positive numbers, so we get

$ \begin{eqnarray*} \frac{x^2 + \frac{4}{x^2}}{2} &\ge& \sqrt{x^2\frac{4}{x^2}} \\ &\ge& \sqrt{4} \\ &\ge& 2 \\ x^2 + \frac{4}{x^2} &\ge& 4 \\ x^2 + 5 + \frac{4}{x^2} &\ge& 9 \\ x^2 + x\frac{1}{x} + x\frac{4}{x} + \frac{4}{x^2} &\ge& 9 \\ (x + \frac{1}{x})(x +\frac{4}{x}) &\ge& 9 \\ \end{eqnarray*} $

There are two interesting facts with the inequality. First, we do not really need to condition $ x > 0 $, it is true all the time. When $ a = b $, the inequality becomes an equality, that means $ x = \pm \sqrt{2} $, indeed if you plug that in, the left hand side has exactly the value 9.