If x + y + z =1 and x, y, z are positive numbers such that (1 – x) (1 – y) (1 – z) `gt=`kxyz, then k =

Options

A. A) 2
B. B) 4
C. C) 8
D. D) 16

Answer

To solve the problem, we need to find the value of \( k \) given the conditions \( x + y + z = 1 \) and \( (1 - x)(1 - y)(1 - z) \geq kxyz \) where \( x, y, z \) are positive numbers.

### Step-by-Step Solution:

1. **Understand the Given Condition:**
We have \( x + y + z = 1 \). This implies that \( 1 - x + 1 - y + 1 - z = 3 - (x + y + z) = 3 - 1 = 2 \).

2. **Express the Product:**
We need to analyze the product \( (1 - x)(1 - y)(1 - z) \). Using the values of \( 1 - x, 1 - y, 1 - z \), we can rewrite it as:
\[ (1 - x)(1 - y)(1 - z) = (1 - x)(1 - y)(1 - z) \]

3. **Apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM):**
By the AM-GM inequality, we know that:
\[ \frac{(1 - x) + (1 - y) + (1 - z)}{3} \geq \sqrt[3]{(1 - x)(1 - y)(1 - z)} \]
Substituting the sum we found:
\[ \frac{2}{3} \geq \sqrt[3]{(1 - x)(1 - y)(1 - z)} \]

4. **Cubing Both Sides:**
Cubing both sides gives:
\[ \left(\frac{2}{3}\right)^3 \geq (1 - x)(1 - y)(1 - z) \]
This simplifies to:
\[ \frac{8}{27} \geq (1 - x)(1 - y)(1 - z) \]

5. **Apply AM-GM to \( x, y, z \):**
Now, applying AM-GM to \( x, y, z \):
\[ \frac{x + y + z}{3} \geq \sqrt[3]{xyz} \]
Since \( x + y + z = 1 \):
\[ \frac{1}{3} \geq \sqrt[3]{xyz} \]
Cubing both sides gives:
\[ \left(\frac{1}{3}\right)^3 \geq xyz \]
Which simplifies to:
\[ \frac{1}{27} \geq xyz \]

6. **Combining the Results:**
We have two inequalities:
- \( (1 - x)(1 - y)(1 - z) \geq kxyz \)
- \( (1 - x)(1 - y)(1 - z) \leq \frac{8}{27} \)
- \( xyz \leq \frac{1}{27} \)

Therefore, we can combine these to find \( k \):
\[ \frac{8}{27} \geq k \cdot \frac{1}{27} \]
Multiplying both sides by 27 gives:
\[ 8 \geq k \]

7. **Conclusion:**
The maximum value of \( k \) that satisfies the inequality is \( k = 8 \).

Thus, the value of \( k \) is \( \boxed{8} \).