How to prove $\frac xy + \frac yx \ge 2$ I am practicing some homework and I'm stumped.
The question asks you to prove that
$x \in Z^+, y \in Z^+$
$\frac xy + \frac yx \ge 2$
So I started by proving that this is true when x and y have the same parity, but I'm not sure how to proceed when x and y have opposite partiy
This is my proof so far for opposite parity
$x,y \in Z^+ $ $|$ $x \gt 0,$ $y \gt 0$. Let x be even $(x=2a, $ $ a \in Z^+)$ and y be odd $(y=2b+1, $ $b \in Z^+)$. Then,
$\frac xy + \frac yx \ge 2$
$\frac {2a}{2b+1} + \frac {2b+1}{2a} \ge 2$
$\frac {2b^2 + 4a^2 + 4a + 1}{2b(2a+1)} \ge 2$
$4b^2 + 4a^2 +4a + 1 \ge 4b(2a+1)$
$4b^2 + 4a^2 + 4a +1 \ge 8ab + 4b$
$4b^2 - 4b + (2a + 1)^2 \ge 8ab$
$(2b-1)^2 + 1 + (2a+1)^2 \ge 8ab$
I feel like this is the not the correct way to go about proving it, but I can't think of a better way to do it. Does anyone have any suggestions? Just hints please, not a solution. 

Without loss of generality, we assume $t:=\frac{x}{y}\geq 1,$ we see that the $t+\frac{1}{t}$ has its derivative non-negative, so $t+\frac{1}{t}\geq 2.$
