If x,y and z are three distinct positive real numbers such that x+y+z=1 then the value of(1//x-1)(1//y-1)(1//z-1)is:

Options

A. 16
B. 8
C. 4
D. 1

Answer

To solve the problem, we need to find the value of \((\frac{1}{x} - 1)(\frac{1}{y} - 1)(\frac{1}{z} - 1)\) given that \(x + y + z = 1\) and \(x, y, z\) are distinct positive real numbers.

### Step-by-Step Solution:

1. **Rewrite the expression**:
We start with the expression:
\[ \left(\frac{1}{x} - 1\right)\left(\frac{1}{y} - 1\right)\left(\frac{1}{z} - 1\right) \]
This can be rewritten as:
\[ \left(\frac{1 - x}{x}\right)\left(\frac{1 - y}{y}\right)\left(\frac{1 - z}{z}\right) \]

2. **Substitute \(1 - x\), \(1 - y\), and \(1 - z\)**:
From the equation \(x + y + z = 1\), we can express:
\[ 1 - x = y + z, \quad 1 - y = x + z, \quad 1 - z = x + y \]
Therefore, we can substitute these into our expression:
\[ \left(\frac{y + z}{x}\right)\left(\frac{x + z}{y}\right)\left(\frac{x + y}{z}\right) \]

3. **Combine the fractions**:
Now we can combine the fractions:
\[ \frac{(y + z)(x + z)(x + y)}{xyz} \]

4. **Expand the numerator**:
We can expand the numerator:
\[ (y + z)(x + z)(x + y) = (y + z)(xy + xz + yz + zy) \]
This will give us a polynomial in \(x, y, z\).

5. **Use the identity**:
We can use the identity \(x + y + z = 1\) to simplify our calculations. However, we can also use the fact that the product of the sums can be expressed in terms of symmetric sums.

6. **Apply AM-GM inequality**:
By applying the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we can show that:
\[ (y + z)(x + z)(x + y) \leq 8xyz \]
This leads us to conclude that:
\[ \frac{(y + z)(x + z)(x + y)}{xyz} \geq 8 \]

7. **Final Value**:
Thus, we find that:
\[ \left(\frac{1}{x} - 1\right)\left(\frac{1}{y} - 1\right)\left(\frac{1}{z} - 1\right) = \frac{(y + z)(x + z)(x + y)}{xyz} = 8 \]

### Conclusion:
The value of \((\frac{1}{x} - 1)(\frac{1}{y} - 1)(\frac{1}{z} - 1)\) is \(8\).