If x, y, z are non-negative real numbers satisfying `x+y+z=1`, then the minimum value of `((1)/(x)+1)((1)/(y)+1)((1)/(z)+1)`, is

Options

A. 8
B. 16
C. 32
D. 64

Explanation

To find the minimum value of the expression \(\left(\frac{1}{x} + 1\right)\left(\frac{1}{y} + 1\right)\left(\frac{1}{z} + 1\right)\) given the constraint \(x + y + z = 1\) where \(x, y, z\) are non-negative real numbers, we can follow these steps:

### Step 1: Rewrite the Expression
We can rewrite the expression as:
\[ \left(\frac{1}{x} + 1\right) = \frac{1 + x}{x} = \frac{1}{x} + 1 \]
Thus, the expression becomes:
\[ \left(\frac{1 + x}{x}\right)\left(\frac{1 + y}{y}\right)\left(\frac{1 + z}{z}\right) \]

### Step 2: Apply the AM-HM Inequality
Using the Arithmetic Mean - Harmonic Mean (AM-HM) inequality, we know:
\[ \frac{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}{3} \geq \frac{3}{x + y + z} \]
Since \(x + y + z = 1\), we have:
\[ \frac{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}{3} \geq 3 \]
This implies:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq 9 \]

### Step 3: Substitute Back into the Expression
Now substituting this back into our expression:
\[ \left(\frac{1}{x} + 1\right)\left(\frac{1}{y} + 1\right)\left(\frac{1}{z} + 1\right) = \left(\frac{1}{x} + 1\right)\left(\frac{1}{y} + 1\right)\left(\frac{1}{z} + 1\right) \]
Expanding this, we have:
\[ \left(\frac{1}{x} + 1\right) = \frac{1 + x}{x}, \quad \left(\frac{1}{y} + 1\right) = \frac{1 + y}{y}, \quad \left(\frac{1}{z} + 1\right) = \frac{1 + z}{z} \]
The product becomes:
\[ \frac{(1 + x)(1 + y)(1 + z)}{xyz} \]

### Step 4: Use the Constraint
Since \(x + y + z = 1\), we can express \(1 + x + y + z = 2\). Therefore:
\[ (1 + x)(1 + y)(1 + z) = 2 + (x + y + z) + (xy + xz + yz) = 2 + 1 + (xy + xz + yz) = 3 + (xy + xz + yz) \]

### Step 5: Apply AM-GM Inequality
Using the AM-GM inequality on \(xyz\):
\[ \frac{x + y + z}{3} \geq \sqrt[3]{xyz} \Rightarrow 1 \geq 3\sqrt[3]{xyz} \Rightarrow xyz \leq \frac{1}{27} \]

### Step 6: Calculate the Minimum Value
Now substituting the minimum values we found:
\[ \left(\frac{1}{x} + 1\right)\left(\frac{1}{y} + 1\right)\left(\frac{1}{z} + 1\right) \geq \frac{(3 + 1)^3}{xyz} \geq \frac{64}{\frac{1}{27}} = 64 \]

### Conclusion
Thus, the minimum value of \(\left(\frac{1}{x} + 1\right)\left(\frac{1}{y} + 1\right)\left(\frac{1}{z} + 1\right)\) is **64**.