If xyz = (1-x)(1-y)(1-z) where 0 ≤ x, y, z ≤ 1 , then the minimum value of x(1-z) + y(1-x) + z(1-y) is:

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 xyz = (1-x)(1-y)(1-z) where 0 ≤ x, y, z ≤ 1 , then the minimum...

xyz = (1-x)(1-y)(1-z)

where

0 \le x, y, z \le 1

, then the minimum value of

x(1-z) + y(1-x) + z(1-y)

is:

Options

A $\frac{3}{2}$
B $\frac{1}{4}$
C $\frac{3}{4}$
D $\frac{1}{2}$

Correct answer

D. $\frac{1}{2}$

Step-by-step solution

Let

E = x(1-z) + y(1-x) + z(1-y) = (x+y+z) - (xy+yz+zx)

. We are given

xyz = 1 - (x+y+z) + (xy+yz+zx) - xyz

. This implies

2xyz = 1 - ((x+y+z) - (xy+yz+zx)) = 1 - E

. Therefore,

E = 1 - 2xyz

. To minimize

E

, we must maximize

xyz

. By AM-GM on

(1-x), (1-y), (1-z)

similar to the previous question, maximum

xyz

occurs when

x=y=z

. If

x=y=z

, then

x^3 = (1-x)^3 \implies x = 1-x \implies x = \frac{1}{2}

. So maximum

xyz = \frac{1}{8}

. Minimum

E = 1 - 2(\frac{1}{8}) = 1 - \frac{1}{4} = \frac{3}{4}

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