How can it be proven that $\frac{x}{y}+\frac{y}{x}\geq2$, with $x$ and $y$ positive? So I realized that I have to prove it with the fact that $(x-y)^2+2xy=x^2+y^2$ 
So $\frac{(x+y)^2}{xy}+2=\frac{x}{y}+\frac{y}{x}$ $\Leftrightarrow$ $\frac{(x+y)^2}{xy}=\frac{x}{y}+\frac{y}{x}-2$ 
Due to the fact that  $(x+y)^2$ is a square, it will be positive 
$x>0$ and $y>0$ so $xy>0$ 
So $\frac{(x+y)^2}{xy}>0$
So $\frac{x}{y}+\frac{y}{x}>2$ 
But the problem is that I have proven that $\frac{x}{y}+\frac{y}{x}>2$, but in the case of $x=y$, it is equal, so it has to be $\geq$...
Could someone help?

With AM-GM $$\frac { x }{ y } +\frac { y }{ x } \geq 2\sqrt { \frac { x }{ y } \frac { y }{ x }  } = 2$$
