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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+(1a)(1b)(1c)1for0a,b,c1
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+(1a)(1b)(1c)[abc+(1a)(1b)(1c)]2(usingCauchysinequality)(1a)(1b)(1c)+abc=1cab+ac+bc+ab1abc+(1a)(1b)(1c)1

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