If `a+b+c=1` and a, b, c are positive real numbers such that
`(1-a)(1-b)(1-c)gelambda" abc, then "lambda`=

Options

A. 2
B. 4
C. 6
D. 8

Explanation

To solve the problem, we need to find the value of \( \lambda \) given the conditions \( a + b + c = 1 \) and \( (1-a)(1-b)(1-c) \geq \lambda abc \) where \( a, b, c \) are positive real numbers.

### Step-by-step Solution:

1. **Use the condition \( a + b + c = 1 \)**:
Since \( a + b + c = 1 \), we can express \( 1 - a \), \( 1 - b \), and \( 1 - c \) as:
\[ 1 - a = b + c, \quad 1 - b = a + c, \quad 1 - c = a + b \]

2. **Substitute into the inequality**:
We need to analyze the expression \( (1-a)(1-b)(1-c) \):
\[ (1-a)(1-b)(1-c) = (b+c)(a+c)(a+b) \]

3. **Apply the AM-GM inequality**:
We can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to each pair:
\[ \frac{a+b}{2} \geq \sqrt{ab} \implies a+b \geq 2\sqrt{ab} \]
\[ \frac{b+c}{2} \geq \sqrt{bc} \implies b+c \geq 2\sqrt{bc} \]
\[ \frac{c+a}{2} \geq \sqrt{ca} \implies c+a \geq 2\sqrt{ca} \]

4. **Multiply the inequalities**:
Multiplying the three inequalities gives:
\[ (a+b)(b+c)(c+a) \geq (2\sqrt{ab})(2\sqrt{bc})(2\sqrt{ca}) = 8\sqrt{(abc)^2} = 8abc \]

5. **Relate it back to the original inequality**:
From our earlier substitution, we have:
\[ (1-a)(1-b)(1-c) = (b+c)(a+c)(a+b) \geq 8abc \]
Thus, we can conclude:
\[ (1-a)(1-b)(1-c) \geq 8abc \]

6. **Identify the value of \( \lambda \)**:
From the inequality \( (1-a)(1-b)(1-c) \geq \lambda abc \), we see that the maximum value of \( \lambda \) that satisfies this inequality is:
\[ \lambda = 8 \]

### Final Answer:
\[ \lambda = 8 \]