Key idea. Put the difference over one positive denominator.

1. The difference between the proposed left and right sides is

   \[\frac{x^2}{y}+\frac{y^2}{x}-x-y=\frac{x^3+y^3-x^2y-xy^2}{xy}.\]

2. Factor the numerator:

   \[\begin{aligned}x^3+y^3-xy(x+y)&=(x+y)(x^2-xy+y^2-xy)\\&=(x+y)(x-y)^2.\end{aligned}\]

3. Therefore the difference is

   \[\frac{(x+y)(x-y)^2}{xy}\ge0,\]

   because \(x+y>0\), \(xy>0\), and the square is non-negative.

4. Equality holds exactly when \(x-y=0\), or \(x=y\).

Remember. Factoring the numerator of a difference can reveal the entire inequality.