MathDB

6

Part of 2018 Czech-Polish-Slovak Junior Match

Problems(1)

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

3/7/2020
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).
inequalitiesalgebra