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|AB/3- 3ab| -3ab<=1 - ab if |A - 3a| <=1 - a , and |B -3b| <= 1 - b and a,b>0

Source: 2021 Irish Mathematical Olympiad P9

May 30, 2021
inequalitiesalgebra

Problem Statement

Suppose the real numbers a,A,b,Ba, A, b, B satisfy the inequalities: ∣A−3a∣≤1−a   ,   ∣B−3b∣≤1−b|A - 3a| \le 1 - a\,\,\, , \,\,\, |B -3b| \le 1 - b, and a,ba, b are positive. Prove that ∣AB3−3ab∣−3ab≤1−ab.\left|\frac{AB}{3}- 3ab\right | - 3ab \le 1 - ab.
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