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2a^2 b^2/(a^5+b^5)+2b^2 c^2/(b^5+c^5)+2c^2 a^2/(c^5+a^5) <= sum (a+b)/(2ab)

Source: Mathematics Regional Olympiad of Mexico Center Zone 2020 P2

November 14, 2021

algebrainequalities

Problem Statement

Let

a

b

and

c

be positive real numbers, prove that

\frac{2a^2 b^2}{a^5+b^5}+\frac{2b^2 c^2}{b^5+c^5}+\frac{2c^2 a^2}{c^5+a^5}\le\frac{a+b}{2ab}+\frac{b+c}{2bc}+\frac{c+a}{2ca}