Find the maximal value of the real number M such that for all positive real numbers a, b, c the following inequality holds: a3+b3+c3−3abc≥M(∣a−b∣3+∣b−c∣3+∣c−a∣3)
Solution
Answer: M=21. The inequality a3+b3+c3−3abc≥M(∣a−b∣3+∣b−c∣3+∣c−a∣3) for M=1/2 implies from 2(a3+b3+c3−3abc)=(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
The value M=1/2 is maximal, because for very large values of a and very small values of b, c, the value of a3+b3+c3−3abc can be very close to a3, and the value of ∣a−b∣3+∣b−c∣3+∣c−a∣3 can be very close to 2a3.
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