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, $abc \neq 0$.

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

Contradiction.