If `xyz = 1` and `x, y, z gt 0` then the minimum value of the expression `(x+2y)(y+2z)(z+2x)` is

Options

A. `81`
B. `9`
C. `18`
D. `27`

Explanation

To find the minimum value of the expression \((x + 2y)(y + 2z)(z + 2x)\) given that \(xyz = 1\) and \(x, y, z > 0\), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Apply AM-GM Inequality:**
We will apply the AM-GM inequality to each of the terms in the expression:
\[ x + 2y \geq 3\sqrt[3]{x \cdot y^2} \]
\[ y + 2z \geq 3\sqrt[3]{y \cdot z^2} \]
\[ z + 2x \geq 3\sqrt[3]{z \cdot x^2} \]

2. **Multiply the Inequalities:**
Now, we multiply the three inequalities:
\[ (x + 2y)(y + 2z)(z + 2x) \geq \left(3\sqrt[3]{x \cdot y^2}\right)\left(3\sqrt[3]{y \cdot z^2}\right)\left(3\sqrt[3]{z \cdot x^2}\right) \]
This simplifies to:
\[ (x + 2y)(y + 2z)(z + 2x) \geq 27 \sqrt[3]{(xyz)^3} \]

3. **Substitute \(xyz = 1\):**
Since we know that \(xyz = 1\):
\[ (x + 2y)(y + 2z)(z + 2x) \geq 27 \sqrt[3]{1^3} = 27 \]

4. **Conclusion:**
Therefore, the minimum value of the expression \((x + 2y)(y + 2z)(z + 2x)\) is \(27\).

### Final Answer:
The minimum value of \((x + 2y)(y + 2z)(z + 2x)\) is **27**.

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