Prove that $(a+b+c)(ab+bc+ca)\geq9abc$ for $a, b, c \geq0.$
2026-03-26 01:16:43.1774487803
Prove that $(a+b+c)(ab+bc+ca)\geq9abc$ for $a, b, c \geq0.$
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By AM-GM $(a+b+c)(ab+ac+bc)\geq3\sqrt[3]{abc}\cdot3\sqrt[3]{a^2b^2c^2}=9abc$