Question

x and y are two +ve numbers suchs that xy =1. Then the minimum value of x + y is

• A. 4
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. 2

Answer 1

The correct option is D 2

$xy=1$
$s=x+\frac{1}{x}$
use AM & GM
$\frac{x+\frac{1}{x}}{2}\ge (x\times \frac{1}{x}{)}^{\frac{1}{2}}$
$x+\frac{1}{x}\ge 2$