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Mongolian Mathematical Olympiad

Algebra Difficulty 4.8 AIME Prove it Mongolia

Let $a, b, c \ge 0$ be non-negative numbers satisfying $a^3 + b^3 + c^3 = abc + 2$ . Prove that
$a^4 + b^4 + c^4 \ge a + b + c.$

Solution

It suffices to prove
$F(a, b, c) = 2(a^4 + b^4 + c^4) - (a + b + c)(a^3 + b^3 + c^3 - abc) \ge 0$
for all $a, b, c \ge 0$ . We may assume $b \ge a$ and $c \ge a$ and then we have
$F(a, b, c) = (b-c)^2(b(b-a) + bc + c(c-a)) + a^2(c-a)(b-a) \ge 0.$
Equality holds for $(a, b, c) = (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 1, 1)$ .

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