Romanian Mathematical Olympiad - District Round

If $a \ge b \ge c \ge 0$, then $d \ge 0$. From $d^2 = \sqrt{ab} + \sqrt{ac} + \sqrt{bc} \le a + b + c \le 2a + 2b + 2c \le 4d$, at least one of the equalities being strict, follows $d \in \{0, 1, 2, 3\}$.

Case I: $d = 0$. Then $a = b = c = 0$.

Case II: $d = 1$. Then $a + b + c = 2$ and $\sqrt{ab} + \sqrt{bc} + \sqrt{ca} = 1$, hence $a = b = 1$, $c = 0$.

Case III: $d \in \{2, 3\}$. When $d = 2$ we have the possibilities $(2, 1, 1)$, $(2, 2, 0)$, $(3, 1, 0)$, $(4, 0, 0)$, and for $d = 3$ we have the possibilities $(3, 2, 1)$, $(3, 3, 0)$, $(4, 2, 0)$, $(4, 1, 1)$, $(5, 1, 0)$, $(6, 0, 0)$; none of them is convenient.

If at least one of the numbers is negative, then all are non-positive, and $-a, -b, -c, -d$ verify the statement.

Finally, we get the solutions $(0, 0, 0)$, $(0, 1, 1)$, $(1, 0, 1)$, $(1, 1, 0)$, $(0, -1, -1)$, $(-1, 0, -1)$, $(-1, -1, 0)$.