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

Note that $y=1/x$ and $$\left(x+\frac{1}{x}\right)^2=\left( x-\frac{1}{x}\right)^2+4\cdot x\cdot \frac{1}{x}\geq4$$ Hence $x+y\geq 2$. Also for $x=y=1$, you have $x+y=2$, hence $2$ indeed the minimum value.
