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

Options

A. `3/2`
B. `1/4`
C. `3/4`
D. `1/2`

Answer

To solve the problem, we need to find the minimum value of the expression \( x(1-z) + y(1-x) + z(1-y) \) given that \( xyz = (1-x)(1-y)(1-z) \) and \( 0 \leq x, y, z \leq 1 \).

### Step-by-Step Solution:

1. **Understand the Given Equation**:
We have the equation:
\[ xyz = (1-x)(1-y)(1-z) \]
This implies a relationship between \( x, y, z \) and their complements \( 1-x, 1-y, 1-z \).

2. **Use the Arithmetic Mean-Geometric Mean Inequality (AM-GM)**:
We can apply the AM-GM inequality to the expression we want to minimize:
\[ \frac{x(1-z) + y(1-x) + z(1-y)}{3} \geq \sqrt[3]{x(1-z) \cdot y(1-x) \cdot z(1-y)} \]

3. **Set Up the Inequality**:
From the AM-GM inequality, we have:
\[ x(1-z) + y(1-x) + z(1-y) \geq 3 \sqrt[3]{x(1-z) \cdot y(1-x) \cdot z(1-y)} \]

4. **Find the Geometric Mean**:
We need to express \( x(1-z) \cdot y(1-x) \cdot z(1-y) \) in a manageable form. Since \( xyz = (1-x)(1-y)(1-z) \), we can substitute values for \( x, y, z \) to find the minimum.

5. **Assume Equal Values**:
To find the minimum, we can assume \( x = y = z \). Let \( x = y = z = k \). Then:
\[ k^3 = (1-k)^3 \]
Taking the cube root gives:
\[ k = 1-k \implies 2k = 1 \implies k = \frac{1}{2} \]

6. **Substitute Back**:
Now substitute \( k = \frac{1}{2} \) into the expression we want to minimize:
\[ x(1-z) + y(1-x) + z(1-y) = \frac{1}{2}(1-\frac{1}{2}) + \frac{1}{2}(1-\frac{1}{2}) + \frac{1}{2}(1-\frac{1}{2} \]
Simplifying this gives:
\[ = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4} \]

7. **Conclusion**:
Therefore, the minimum value of \( x(1-z) + y(1-x) + z(1-y) \) is:
\[ \boxed{\frac{3}{4}} \]