How to prove $\frac xy + \frac yx \ge 2$

Clearly $(x-y)^2 \geq 0$

So $x^2-2xy+y^2 \geq 0 \Rightarrow x^2+y^2 \geq 2xy $

Since $x $ and $y$ are positive integers , $x \neq 0 $ and $y \neq 0$.

Thus we can divide by $xy$.

So we have $\frac{x^2+y^2}{xy} \geq \frac {2xy}{xy} \Rightarrow \frac {x}{y}+\frac{y}{x} \geq 2$.

Hint: Consider multiplying both sides of the inequality by $xy$, which is positive. Then, we have: $$\frac xy + \frac yx \ge 2\\ x^2+y^2\ge 2xy $$ Can you go on?

$$(x-y)^2\geq0$$

$$\begin{array}{l}x^2-2xy+y^2\geq0\\x^2+y^2\geq2xy\\\frac{x^2+y^2}{xy}\geq2\\\frac{x^2}{xy}+\frac{y^2}{xy}\geq2\\\frac{x}{y}+\frac{y}{x}\geq2\\QED\end{array}$$

This will work for $x,y\in\mathbb{R}\backslash\{0\}$