Ukrainian National Mathematical Olympiad, Third Round, First Tour

If $x = 0$ or $y = 0$, the inequality is trivial. Now assume $x, y > 0$. Divide both parts of the inequality by $xy$. We get:
$$xy + x + y \leq x^3 + \frac{1}{y} + \frac{y^3}{x}.$$
Apply the famous inequality, which is a consequence of the Cauchy-Schwarz inequality:
$$\frac{a_1^2}{b_1} + \frac{a_2^2}{b_2} + \cdots + \frac{a_n^2}{b_n} \geq \frac{(a_1 + a_2 + \cdots + a_n)^2}{b_1 + b_2 + \cdots + b_n}$$
$$x^3 + \frac{1}{y} + \frac{y^3}{x} \geq \frac{(x^2)^2}{x} + \frac{1^2}{y} + \frac{(y^2)^2}{xy} \geq \frac{(x^2 + 1 + y^2)^2}{x + y + xy} \geq \frac{(x + y + xy)^2}{x + y + xy} = xy + x + y,$$
which completes the proof.