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Black Advanced Problems in Mathematics
Exam
Black Advanced Problems in Mathematics
Subject
Mathematics
Chapter
Sequences and Series

Black Advanced Problems in Mathematics · Mathematics · Sequences and Series

If a, b, c are positive numbers and a + b + c = 1 , then the...

If a,b,ca, b, c are positive numbers and a+b+c=1a + b + c = 1, then the maximum value of (1−a)(1−b)(1−c)(1 - a)(1 - b)(1 - c) is:

Options

  1. A11
  2. B23\frac{2}{3}
  3. C827\frac{8}{27}
  4. D49\frac{4}{9}

Correct answer

D. 49\frac{4}{9}

Step-by-step solution

We know that a+b+c=1a+b+c=1. Then 1−a=b+c1-a = b+c, 1−b=a+c1-b = a+c, and 1−c=a+b1-c = a+b. We want to maximize (b+c)(a+c)(a+b)(b+c)(a+c)(a+b). By AM-GM inequality: b+c2≥bca+c2≥aca+b2≥ab\frac{b+c}{2} \ge \sqrt{bc}\frac{a+c}{2} \ge \sqrt{ac}\frac{a+b}{2} \ge \sqrt{ab} Multiplying them gives (b+c)(a+c)(a+b)≥8abc(b+c)(a+c)(a+b) \ge 8abc. Alternatively, using AM-GM on (1−a),(1−b),(1−c)(1-a), (1-b), (1-c): (1−a)+(1−b)+(1−c)3≥(1−a)(1−b)(1−c)33−(a+b+c)3≥(1−a)(1−b)(1−c)323≥(1−a)(1−b)(1−c)3\frac{(1-a)+(1-b)+(1-c)}{3} \ge \sqrt[3]{(1-a)(1-b)(1-c)}\frac{3-(a+b+c)}{3} \ge \sqrt[3]{(1-a)(1-b)(1-c)}\frac{2}{3} \ge \sqrt[3]{(1-a)(1-b)(1-c)} Taking the cube on both sides: (1−a)(1−b)(1−c)≤827(1-a)(1-b)(1-c) \le \frac{8}{27}
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