Given that `x,y,z` are positive real numbers such that `xyz=32`, the minimum value of `sqrt((x+2y)^(2)+2z^(2)-15)` is equal to

Options

A. `6`
B. `8`
C. `9`
D. `12`

Explanation

To find the minimum value of \( \sqrt{(x + 2y)^2 + 2z^2 - 15} \) given that \( xyz = 32 \) and \( x, y, z \) are positive real numbers, we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression inside the square root:
\[ (x + 2y)^2 + 2z^2 - 15 \]
We will analyze \( (x + 2y)^2 + 2z^2 \).

### Step 2: Apply the AM-GM inequality
Using the AM-GM inequality, we can find a lower bound for the expression \( (x + 2y)^2 + 2z^2 \). According to AM-GM:
\[ \frac{(x + 2y)^2 + 2z^2}{3} \geq \sqrt[3]{(x + 2y)^2 \cdot 2z^2} \]
This implies:
\[ (x + 2y)^2 + 2z^2 \geq 3 \sqrt[3]{(x + 2y)^2 \cdot 2z^2} \]

### Step 3: Substitute \( z \) using \( xyz = 32 \)
From the condition \( xyz = 32 \), we can express \( z \) as:
\[ z = \frac{32}{xy} \]
Substituting this into our expression gives us:
\[ 2z^2 = 2 \left( \frac{32}{xy} \right)^2 = \frac{2048}{x^2y^2} \]

### Step 4: Simplify the expression
Now we need to minimize:
\[ (x + 2y)^2 + \frac{2048}{x^2y^2} \]
Let \( a = x + 2y \). We can express \( a^2 + \frac{2048}{x^2y^2} \) in terms of \( a \) and \( xy \).

### Step 5: Use AM-GM again
We can apply AM-GM again on \( a^2 \) and \( \frac{2048}{x^2y^2} \):
\[ \frac{a^2 + \frac{2048}{x^2y^2}}{2} \geq \sqrt{a^2 \cdot \frac{2048}{x^2y^2}} \]
This gives us:
\[ a^2 + \frac{2048}{x^2y^2} \geq 2 \sqrt{a^2 \cdot \frac{2048}{x^2y^2}} \]

### Step 6: Find a specific case
To find the minimum, we can assume \( x = 4 \), \( y = 2 \), and \( z = 4 \) (since \( 4 \cdot 2 \cdot 4 = 32 \)):
\[ (x + 2y) = 4 + 4 = 8 \]
\[ 2z^2 = 2 \cdot 4^2 = 32 \]
Thus:
\[ (x + 2y)^2 + 2z^2 = 8^2 + 32 = 64 + 32 = 96 \]

### Step 7: Calculate the minimum value
Now we substitute back into our original expression:
\[ \sqrt{(x + 2y)^2 + 2z^2 - 15} = \sqrt{96 - 15} = \sqrt{81} = 9 \]

### Conclusion
Thus, the minimum value of \( \sqrt{(x + 2y)^2 + 2z^2 - 15} \) is:
\[ \boxed{9} \]