If a,b,c are all positive then the minimum value of the expression`(a^2+a+1)(b^2+b+1)(c^2+c+1)//abc` is:

Options

A. 3
B. 9
C. 27
D. 1

Answer

To find the minimum value of the expression \((a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1) / (abc)\) where \(a, b, c\) are all positive, we can follow these steps:

### Step 1: Analyze the individual terms
We start by examining the term \(f(x) = \frac{x^2 + x + 1}{x}\).

### Step 2: Simplify \(f(x)\)
We can rewrite \(f(x)\) as:
\[ f(x) = x + \frac{1}{x} + 1 \]

### Step 3: Apply the AM-GM inequality
Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we know that:
\[ x + \frac{1}{x} \geq 2 \]
for \(x > 0\). Thus,
\[ f(x) = x + \frac{1}{x} + 1 \geq 2 + 1 = 3 \]

### Step 4: Apply the result to \(a\), \(b\), and \(c\)
Since \(f(a) \geq 3\), \(f(b) \geq 3\), and \(f(c) \geq 3\), we can multiply these inequalities:
\[ f(a) \cdot f(b) \cdot f(c) \geq 3 \cdot 3 \cdot 3 = 27 \]

### Step 5: Substitute back into the original expression
Now substituting back into the original expression, we have:
\[ \frac{(a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1)}{abc} \geq \frac{27}{abc} \]

### Step 6: Find the minimum value
To achieve the minimum value of the expression, we need \(abc\) to be as small as possible while keeping \(a\), \(b\), and \(c\) positive. The minimum occurs when \(a = b = c = 1\):
\[ \frac{(1^2 + 1 + 1)(1^2 + 1 + 1)(1^2 + 1 + 1)}{1 \cdot 1 \cdot 1} = \frac{3 \cdot 3 \cdot 3}{1} = 27 \]

### Conclusion
Thus, the minimum value of the expression is:
\[ \boxed{27} \]