mathschallenge.net

Problem

Consider the function, $f(x) = x + \dfrac{1}{x}$.

$f(2) = 2 + \dfrac{1}{2} = 2.5$
$f(0.4) = 0.4 + \dfrac{1}{0.4} = 2.9$
$f(1.25) = 1.25 + \dfrac{1}{1.25} = 2.05$

Prove that $f(x) \ge 2$ for all real values of $x \gt 0$.

Solution

Clearly $(x - 1)^2 \ge 0$ for all real values of $x$.

$\begin {align}\therefore x^2 - 2x + 1 &\ge 0\\x^2 + 1 &\ge 2x\end {align}$

As $x \gt 0$ we can divide through by $x$ without changing the sign of the inequality.

$\therefore x + \dfrac{1}{x} \ge 2$       Q.E.D.

1. Prove that $\dfrac{(x + 1)^2}{x} \ge 4$ for $x \gt 0$.
2. Determine the domain of $x$ for which $x^2 \ge 4(x - 1)$ is true.