minimum sum 2

Let $x$ be the number and $s$ be the sum, then $s=x+\dfrac{1}{x}$ .

Differentiating both sides with respect to $x$ , we have $\dfrac{ds}{dx}=1-\dfrac{1}{x^2}$ .

For the sum to be minimum, $\dfrac{ds}{dx}=0$ . We have

$\dfrac{1}{x^2}=1$ $\implies$ $x^2=1$ $\implies$ $x \pm 1$

Since $x$ is a positive number, $x=1$ .

By the second derivative check, we have

$\dfrac{d^2s}{dx^2}=\dfrac{2}{x^3}$

If $x=1$ , then

$\dfrac{d^2s}{dx^2}=2$

Since $2>0$ , $s$ is a minimum.

By AM-GM Inequality , $a+\frac 1a \ge 2 \sqrt{a \cdot \frac 1a} = 2$ Equality occurs at $\boxed{a=1}$