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a^3 + b^3 = 2 => 1/a+ 1/b \ge 2(a^2 - a + 1)(b^2 - b + 1) for a,b>0

Source: Czech-Polish-Slovak Junior Match 2018, Team p6 CPSJ

March 7, 2020
inequalitiesalgebra

Problem Statement

Positive real numbers a,ba, b are such that a3+b3=2a^3 + b^3 = 2. Show that that 1a+1b≥2(a2−a+1)(b2−b+1)\frac{1}{a}+\frac{1}{b}\ge 2(a^2 - a + 1)(b^2 - b + 1).
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