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Question Number 63247 by Joel122 last updated on 01/Jul/19

Prove that (√(abc)) + (√((1−a)(1−b)(1−c))) ≤ 1 for 0 ≤ a,b,c ≤ 1

Provethatabc+(1−a)(1−b)(1−c)⩽1for0⩽a,b,c⩽1

Commented by Tony Lin last updated on 01/Jul/19

∵ abc+(1−a)(1−b)(1−c)≥ [(√(abc))+(√((1−a)(1−b)(1−c)))]^2 (using Cauchy′s inequality) (1−a)(1−b)(1−c)+abc =1−c−a−b+ac+bc+ab≤1 ∴ (√(abc))+(√((1−a)(1−b)(1−c)))≤1

∵abc+(1−a)(1−b)(1−c)⩾[abc+(1−a)(1−b)(1−c)]2(usingCauchy′sinequality)(1−a)(1−b)(1−c)+abc=1−c−a−b+ac+bc+ab⩽1∴abc+(1−a)(1−b)(1−c)⩽1

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