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A funny inequality with (possibly negative) real numbers

Source: 2024 Polish Junior Math Olympiad Finals P3

May 5, 2024
inequalitiesinequalities proposedalgebraalgebra proposed

Problem Statement

Real numbers a,b,ca,b,c satisfy a+b≠0a+b \ne 0, b+c≠0b+c \ne 0 and c+a≠0c+a \ne 0. Show that (a2ca+b+b2ab+c+c2bc+a)⋅(b2ca+b+c2ab+c+a2bc+a)≥0.\left(\frac{a^2c}{a+b}+\frac{b^2a}{b+c}+\frac{c^2b}{c+a}\right) \cdot \left(\frac{b^2c}{a+b}+\frac{c^2a}{b+c}+\frac{a^2b}{c+a}\right) \ge 0.
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