If a, b, c are positive real numbers such that `a+b+c=1`, then the greatest value of `(1-a)(1-b)(1-c), is

Options

A. A) 1
B. B) `2/3`
C. C) `8/27`
D. D) `4/9`

Answer

To find the greatest value of \((1-a)(1-b)(1-c)\) given that \(a + b + c = 1\) and \(a, b, c\) are positive real numbers, we can use the method of inequalities, specifically the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Express the terms**:
We know that \(1 - a = b + c\), \(1 - b = a + c\), and \(1 - c = a + b\). Therefore, we can rewrite the expression we want to maximize:
\[ (1-a)(1-b)(1-c) = (b+c)(a+c)(a+b) \]

2. **Apply AM-GM Inequality**:
By the AM-GM inequality, we have:
\[ \frac{(1-a) + (1-b) + (1-c)}{3} \geq \sqrt[3]{(1-a)(1-b)(1-c)} \]
Substituting \(1-a + 1-b + 1-c = 3 - (a+b+c) = 3 - 1 = 2\):
\[ \frac{2}{3} \geq \sqrt[3]{(1-a)(1-b)(1-c)} \]

3. **Cube both sides**:
To eliminate the cube root, we cube both sides:
\[ \left(\frac{2}{3}\right)^3 \geq (1-a)(1-b)(1-c) \]
Simplifying the left side:
\[ \frac{8}{27} \geq (1-a)(1-b)(1-c) \]

4. **Conclusion**:
The maximum value of \((1-a)(1-b)(1-c)\) is \(\frac{8}{27}\).

5. **Check for equality condition**:
The equality in AM-GM holds when all the terms are equal, i.e., \(1-a = 1-b = 1-c\). This implies \(a = b = c = \frac{1}{3}\).

Thus, the greatest value of \((1-a)(1-b)(1-c)\) is \(\frac{8}{27}\).

### Final Answer:
The greatest value of \((1-a)(1-b)(1-c)\) is \(\frac{8}{27}\).