MathDB
2010 BAMO C a+b+c \ge \frac{|a|+|b|+|c|}{3}

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August 27, 2019
algebrainequalities

Problem Statement

Suppose a,b,ca,b,c are real numbers such that a+b≥0,b+c≥0a+b \ge 0, b+c \ge 0, and c+a≥0c+a \ge 0. Prove that a+b+c≥∣a∣+∣b∣+∣c∣3a+b+c \ge \frac{|a|+|b|+|c|}{3} .
(Note: ∣x∣|x| is called the absolute value of xx and is defined as follows. If x≥0x \ge 0 then ∣x∣=x|x|= x, and if x<0x < 0 then ∣x∣=−x|x| = -x. For example, ∣6∣=6,∣0∣=0|6|= 6, |0| = 0 and ∣−6∣=6|-6| = 6.)
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