If a,b,c are non-zero real numbers, then the minimum value of the expression `((a^(8)+4a^(4)+1)(b^(4)+3b^(2)+1)(c^(2)+2c+2))/(a^(4)b^(2))` equals

Options

A. 12
B. 24
C. 30
D. 60

Explanation

To find the minimum value of the expression

\[ \frac{(a^8 + 4a^4 + 1)(b^4 + 3b^2 + 1)(c^2 + 2c + 2)}{a^4 b^2}, \]

we will analyze each part of the expression separately and apply the AM-GM inequality.

### Step 1: Analyze \(a^8 + 4a^4 + 1\)

Let \(x = a^4\). Then we can rewrite the expression as:

\[ a^8 + 4a^4 + 1 = x^2 + 4x + 1. \]

To find the minimum value of \(x^2 + 4x + 1\), we can complete the square:

\[ x^2 + 4x + 1 = (x + 2)^2 - 4 + 1 = (x + 2)^2 - 3. \]

The minimum value occurs when \((x + 2)^2 = 0\), which gives \(x = -2\). However, since \(x = a^4\) must be non-negative, we evaluate at \(x = 0\) (not valid) and find the minimum at \(x = 0\) is not applicable.

Instead, we can apply AM-GM directly:

\[ a^8 + 4a^4 + 1 \geq 6 \sqrt[6]{a^8 \cdot a^4 \cdot a^4 \cdot a^4 \cdot 1 \cdot 1} = 6. \]

Thus,

\[ \frac{(a^8 + 4a^4 + 1)}{a^4} \geq 6. \]

### Step 2: Analyze \(b^4 + 3b^2 + 1\)

Let \(y = b^2\). Then we can rewrite the expression as:

\[ b^4 + 3b^2 + 1 = y^2 + 3y + 1. \]

Completing the square gives:

\[ y^2 + 3y + 1 = \left(y + \frac{3}{2}\right)^2 - \frac{9}{4} + 1 = \left(y + \frac{3}{2}\right)^2 - \frac{5}{4}. \]

The minimum value occurs at \(y = -\frac{3}{2}\) (not valid), thus we apply AM-GM:

\[ b^4 + 3b^2 + 1 \geq 5 \sqrt[5]{b^4 \cdot b^2 \cdot 1 \cdot 1 \cdot 1} = 5. \]

Thus,

\[ \frac{(b^4 + 3b^2 + 1)}{b^2} \geq 5. \]

### Step 3: Analyze \(c^2 + 2c + 2\)

Completing the square gives:

\[ c^2 + 2c + 2 = (c + 1)^2 + 1 \geq 1. \]

### Step 4: Combine the Results

Now we combine the results:

\[ \frac{(a^8 + 4a^4 + 1)(b^4 + 3b^2 + 1)(c^2 + 2c + 2)}{a^4 b^2} \geq \frac{6 \cdot 5 \cdot 1}{1} = 30. \]

### Conclusion

Thus, the minimum value of the expression is

\[ \boxed{30}. \]