If `x,y,x gt 0 and x+y+z=1` then `(xyz)/((1-x) (1-y) (1-z))` is necessarily.

Options

A. `ge8`
B. `le 1/8`
C. `1`
D. None of these

Answer

To solve the problem, we need to find the value of \(\frac{xyz}{(1-x)(1-y)(1-z)}\) given that \(x, y, z > 0\) and \(x + y + z = 1\).

### Step-by-Step Solution:

1. **Using AM-GM Inequality for \(xyz\)**:
Since \(x, y, z\) are positive and \(x + y + z = 1\), we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{x + y + z}{3} \geq \sqrt[3]{xyz} \]
Substituting \(x + y + z = 1\):
\[ \frac{1}{3} \geq \sqrt[3]{xyz} \]
Cubing both sides (since all variables are positive):
\[ \left(\frac{1}{3}\right)^3 \geq xyz \implies \frac{1}{27} \geq xyz \]
Thus, we have:
\[ xyz \leq \frac{1}{27} \quad \text{(Equation 1)} \]

2. **Using AM-GM Inequality for \((1-x)(1-y)(1-z)\)**:
We rewrite the terms \(1-x\), \(1-y\), and \(1-z\):
\[ 1 - x = y + z, \quad 1 - y = x + z, \quad 1 - z = x + y \]
Now, applying AM-GM again:
\[ \frac{(1-x) + (1-y) + (1-z)}{3} \geq \sqrt[3]{(1-x)(1-y)(1-z)} \]
The left-hand side simplifies to:
\[ \frac{3 - (x + y + z)}{3} = \frac{3 - 1}{3} = \frac{2}{3} \]
Therefore:
\[ \frac{2}{3} \geq \sqrt[3]{(1-x)(1-y)(1-z)} \]
Cubing both sides gives:
\[ \left(\frac{2}{3}\right)^3 \geq (1-x)(1-y)(1-z) \implies \frac{8}{27} \geq (1-x)(1-y)(1-z) \quad \text{(Equation 2)} \]

3. **Finding the Ratio**:
We need to find:
\[ \frac{xyz}{(1-x)(1-y)(1-z)} \]
Using the inequalities from Equations 1 and 2:
\[ \frac{xyz}{(1-x)(1-y)(1-z)} \leq \frac{\frac{1}{27}}{\frac{8}{27}} = \frac{1}{8} \]

Thus, we conclude that:
\[ \frac{xyz}{(1-x)(1-y)(1-z)} \leq \frac{1}{8} \]

### Final Answer:
The expression \(\frac{xyz}{(1-x)(1-y)(1-z)}\) is necessarily less than or equal to \(\frac{1}{8}\).