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Silk Road Mathematics Competition competition problems

Algebra Difficulty 4.9 AIME Prove it Silk Road Mathematics Competition

Find the maximal value of the real number MM such that for all positive real numbers aa, bb, cc the following inequality holds:
a3+b3+c3−3abc≥M(∣a−b∣3+∣b−c∣3+∣c−a∣3) a^3 + b^3 + c^3 - 3abc \ge M(|a-b|^3 + |b-c|^3 + |c-a|^3)

Solution

Answer: M=12M = \frac{1}{2}.
The inequality a3+b3+c3−3abc≥M(∣a−b∣3+∣b−c∣3+∣c−a∣3)a^3 + b^3 + c^3 - 3abc \ge M(|a-b|^3 + |b-c|^3 + |c-a|^3) for M=1/2M = 1/2 implies from
2(a3+b3+c3−3abc)=(a+b+c)((a−b)2+(b−c)2+(c−a)2) 2(a^3 + b^3 + c^3 - 3abc) = (a + b + c)((a - b)^2 + (b - c)^2 + (c - a)^2)
and from obvious inequalities
(x+y+z)(x−y)2≥∣x−y∣3 (x + y + z)(x - y)^2 \ge |x - y|^3

The value M=1/2M = 1/2 is maximal, because for very large values of aa and very small values of bb, cc, the value of a3+b3+c3−3abca^3 + b^3 + c^3 - 3abc can be very close to a3a^3, and the value of ∣a−b∣3+∣b−c∣3+∣c−a∣3|a-b|^3 + |b-c|^3 + |c-a|^3 can be very close to 2a32a^3.

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