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1+2abc>=a^2+b^2+c^2 => 1+2(abc)^2>=a^4+b^4+c^4 (HOMC 2015 J Q6)

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August 6, 2019
algebrainequalities

Problem Statement

Let a,b,c[1,1]a, b, c \in [-1, 1] such that 1+2abca2+b2+c21 + 2abc \ge a^2 + b^2 + c^2. Prove that 1+2a2b2c2a4+b4+c41 + 2a^2b^2c^2 \ge a^4 + b^4 + c^4.