$x$ and $y$ be two variables such that $x > 0$ and $xy = 1.$ Then the minimum value of $x + y$ is: 

$xy=1\Rightarrow y=\frac{1}{x}\text{ and let }z=x+y$

$z=x+\frac{1}{x}\Rightarrow \frac{dz}{dx}=1-\frac{1}{x^2}$

$\Rightarrow \frac{dz}{dx}=0\Rightarrow 1-\frac{1}{x^2}=0$

$\Rightarrow x=-1,+1\text{ and }\frac{d^{2}z}{d{x}^2}=\frac{2}{x^3}$

    ${\left(\frac{d^{2}z}{d{x}^2}\right)}_{x=1}=\frac{2}{1}=2=+ve$

$\therefore x=1$ is point of minima. $x=1, y=1,$

$\therefore$ minimum value$=x+y=2.$