Problem 681

*First solution.* If $a + b + c \le 0$, the inequality is obviously satisfied, with equality occurring if and only if $a = b = c = 0$. If $a + b + c > 0$, then $abc < 0$. It is not possible for all the variables to be negative (their sum would be negative), therefore one of them is negative and the other two are positive. We may assume that $a < 0 < b, c$.

Put $x = -a$. Then $-x + b + c = xbc > 0$ and $(a^2 + b^2 + c^2)^2 = (x^2 + b^2 + c^2)^2 \ge (xb + xc + bc)^2 \ge 3(xb \cdot xc + xc \cdot bc + bc \cdot xb) = 3xbc(x + b + c) = 3(-x + b + c)(x + b + c) \ge 9(-x + b + c)^2$.
As $-x + b + c > 0$, the last inequality comes to $x + b + c \ge 3(-x + b + c)$, i.e. to $x \ge \frac{b + c}{2}$. But $x = \frac{b + c}{1 + bc} \ge \frac{b + c}{2}$ because $bc \le 1$. In conclusion, we obtain $a^2 + b^2 + c^2 \ge 3(-x + b + c) = 3(a + b + c)$, with equality (in the case $a < 0 < b, c$) if and only if $x = b = c$ and $bc = 1$, i.e., if $a = -1$, $b = c = 1$. To conclude, we have equality if $(a, b, c) \in \{(0, 0, 0), (-1, 1, 1), (1, -1, 1), (1, 1, -1)\}$.

*Second solution.* If $a + b + c \le 0$, the inequality is obviously satisfied, with equality occurring if and only if $a = b = c = 0$. If $a + b + c > 0$, then $abc < 0$.
From the AM-GM inequality, $a^2 + b^2 + c^2 \ge 3\sqrt[3]{(abc)^2} \ge 3|abc| = -3abc = 3(a + b + c)$, with equality if $a^2 = b^2 = c^2$, $abc \le 0$ and $|abc| = 1$, which leads to $(a, b, c) \in \{(0, 0, 0), (-1, 1, 1), (1, -1, 1), (1, 1, -1)\}$.