62nd 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: $$xy+x+y\le x^3+\frac{1}{y}+\frac{y^3}{x}.$$ Apply the famous inequality, which is a consequence of the Cauchy-Schwarz inequality: $$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\cdots +\frac{a_n^2}{b_n}\ge \frac{(a_1+a_2+\cdots +a_n{)}^2}{b_1+b_2+\cdots +b_n}$$ $$x^3+\frac{1}{y}+\frac{y^3}{x}\ge \frac{(x^2{)}^2}{x}+\frac{1^2}{y}+\frac{(y^2{)}^2}{xy}\ge \frac{(x^2+1+y^2{)}^2}{x+y+xy}\ge \frac{(x+y+xy{)}^2}{x+y+xy}=xy+x+y,$$ which completes the proof.