$xy=1 \implies $minimum $x+y=$? If $x,y$ are real positive numbers such that $xy=1$, how can I find the minimum for $x+y$?

$x+y=x+y-2\sqrt{xy}+2\sqrt{xy}=(\sqrt{x}-\sqrt{y})^2+2\ge2$
The value will be attained when $(\sqrt{x}-\sqrt{y})^2=0\Rightarrow \sqrt{x}=\sqrt{y}\Rightarrow x=y=1$
Please note that this is the basis for the inequality of A.M.$\ge$ G.M.
If $x,y\ge 0$ then we know that $(\sqrt{x}-\sqrt{y})^2\ge0$
$\Rightarrow x+y-2\sqrt{xy}\ge 0$
$\Rightarrow x+y\ge 2\sqrt{xy}$
