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xyz(x+y+z)+2021 >= 2024xyz

Source: Cyprus 2021Junior TST-2 Problem 1

May 26, 2021
algebrainequalitiesAM-GM

Problem Statement

Let x,y,zx,y,z be positive real numbers such that x2+y2+z2=3x^2+y^2+z^2=3. Prove that xyz(x+y+z)+2021⩾2024xyz xyz(x+y+z)+2021\geqslant 2024xyz
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