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

If for  $c>0, xy=c^2\implies y=\frac{c^2}x\implies x+y=x+\frac{c^2}x=z$(say)
So, $x^2-zx+c^2=0$ 
As this is a quadratic equation in $x$ and $x$ is real,
the discriminant  $(-z)^2-4\cdot1\cdot c^2\ge 0\implies z^2\ge4c^2\implies $ either $z\ge2c$ or $z\le -2c$
As $x,y>0,x+y>0$ 
So,  $x+y=x+\frac1x=z\ge2c$
