Romanian Mathematical Olympiad - Final Round

By adding $2ab + 2bc + 2ca$ to both sides, the inequality becomes:
$$(a + b + c)^2 + a^2b + b^2c + c^2a \ge 6 + 2ab + 2bc + 2ca.$$
Thus, we have to prove that $a^2b + b^2c + c^2a + 3 \ge 2ab + 2bc + 2ca$. Since $a + b + c = 3$, the previous inequality is equivalent to:
$$(b + a^2b) + (c + b^2c) + (a + c^2a) \ge 2ab + 2bc + 2ca. \quad (1) *Alternative solution.* By adding \$a + b + c\$ to both sides, the inequality becomes:$$
a^2 + b^2 + c^2 + a^2b + b^2c + c^2a + a + b + c $\ge 9. \quad$ (2)
$$$$
Using the obvious inequalities $b + a^2b \ge 2ab$, $c + b^2c \ge 2bc$ and $a + c^2a \ge 2ca$, we deduce: $a^2 + b^2 + c^2 + a^2b + b^2c + c^2a + a + b + c \ge a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2 = 9$, therefore (2) is true, which ends the proof.

It is obvious that $b + a^2b \ge 2ab$, $c + b^2c \ge 2bc$ and $a + c^2a \ge 2ca$. By summing these inequalities, we find that (1) is true, which ends the proof.