58th Ukrainian National Mathematical Olympiad

We consider equality from the problem as a quadratic equation with respect to the variable $c$: $$c^2+abc+(a^2+b^2-4)=0\Rightarrow c=\frac{-ab\pm \sqrt{a^2{b}^2+16-4a^2-4b^2}}{2}.$$ Since $c$ is a positive number, then $c=\frac{-ab+\sqrt{a^2{b}^2+16-4a^2-4b^2}}{2}$. Now we can solve the problem in the following way: $$\begin{gathered} \frac{-ab+\sqrt{a^2{b}^2+16-4a^2-4b^2}}{2}+ab\le 2\iff \sqrt{a^2{b}^2+16-4a^2-4b^2}\le 4-ab\iff \\ {a}^2{b}^2+16-4a^2-4b^2\le 16-8ab+a^2{b}^2\iff 4a^2+4b^2\ge 8ab\iff 4(a-b{)}^2\ge 0. \end{gathered}$$

Note that $ab\le 4$, because otherwise $a^2+b^2\ge 2ab\ge 8$, and this contradicts to the problem statement.