The minimum value of `((a^2 +3a+1)(b^2+3b + 1)(c^2+ 3c+ 1))/(abc)`The minimum value of , where `a, b, c in R^+` is

Options

A. `(11^(3))/(2^(3))`
B. 125
C. 25
D. 27

Explanation

To find the minimum value of the expression \(\frac{(a^2 + 3a + 1)(b^2 + 3b + 1)(c^2 + 3c + 1)}{abc}\) where \(a, b, c \in \mathbb{R}^+\), we can follow these steps:

### Step 1: Rewrite the expression
We start by rewriting the expression:
\[ \frac{(a^2 + 3a + 1)(b^2 + 3b + 1)(c^2 + 3c + 1)}{abc} \]
We can separate each term in the numerator:
\[ \frac{(a^2 + 3a + 1)}{a} \cdot \frac{(b^2 + 3b + 1)}{b} \cdot \frac{(c^2 + 3c + 1)}{c} \]

### Step 2: Simplify each term
Now, simplify each term:
\[ \frac{a^2 + 3a + 1}{a} = a + 3 + \frac{1}{a} \]
Thus, the expression becomes:
\[ \left(a + 3 + \frac{1}{a}\right) \cdot \left(b + 3 + \frac{1}{b}\right) \cdot \left(c + 3 + \frac{1}{c}\right) \]

### Step 3: Find the minimum value of each term
To find the minimum value of \(a + 3 + \frac{1}{a}\), we can use calculus or the AM-GM inequality. Using AM-GM:
\[ a + \frac{1}{a} \geq 2 \quad \text{(with equality when \(a = 1\))} \]
Thus:
\[ a + 3 + \frac{1}{a} \geq 2 + 3 = 5 \]
The minimum value occurs when \(a = 1\).

### Step 4: Apply the same logic for \(b\) and \(c\)
By the same reasoning, we have:
\[ b + 3 + \frac{1}{b} \geq 5 \quad \text{(minimum at \(b = 1\))} \]
\[ c + 3 + \frac{1}{c} \geq 5 \quad \text{(minimum at \(c = 1\))} \]

### Step 5: Combine the results
Thus, the minimum value of the entire expression is:
\[ 5 \cdot 5 \cdot 5 = 125 \]

### Conclusion
The minimum value of \(\frac{(a^2 + 3a + 1)(b^2 + 3b + 1)(c^2 + 3c + 1)}{abc}\) is \(125\).