58th Ukrainian National Mathematical Olympiad

By contradiction, assume that $a+b+c=0$. Thus, $$ab+bc+ca=a+b+c=0.$$

Suppose that $abc=0$, so without loss of generality let $c=0$. Hence $ab=0$. Therefore, two variables are zeros. By the first condition on numbers in the problem, we obtain a contradiction. Therefore, $abc\ne 0$.

Plug $c=-a-b$ in equation $ab+bc+ca=0$. Hence $$\begin{array}{rl}(a+b)b+(a+b)a-ab& =0\\ a^2+ab+b^2& =0\\ a^2+ab+\frac{1}{4}{b}^2+\frac{3}{4}{b}^2& =0\\ (a+\frac{1}{2}b{)}^2+\frac{3}{4}{b}^2& =0\\ a=b=0.& \end{array}$$

Contradiction.