59th Ukrainian National Mathematical Olympiad

Without loss of generality, let $x,y$ be positive. Then we can factor the expression as $(x^4+y^4)(x^2+y^2)$. Since $(x^2-y^2{)}^2=x^4-2x^2{y}^2+y^4\ge 0$, then $x^4+y^4\ge 2x^2{y}^2=2$, and similarly $x^2+y^2\ge 2xy=2$. Therefore, the smallest number is $4$, that can be obtained when $x=y=1$.