Consider the expression
`((a^2 + a + 1)(b^2 + b + 1) (c^2 + c + 1)(d^2 + d+ 1)(e^2 + e + 1))/(abcde)`
where a, b, c, d and e are positive numbers. The minimum value of the expression is

Options

A. A) 3
B. B) 243
C. C) 10
D. D) 100

Answer

To find the minimum value of the expression

\[ \frac{(a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1)(d^2 + d + 1)(e^2 + e + 1)}{abcde} \]

where \(a, b, c, d,\) and \(e\) are positive numbers, we can follow these steps:

### Step 1: Analyze the individual terms
Each term in the numerator can be expressed as:

\[ x^2 + x + 1 \]

for each variable \(x \in \{a, b, c, d, e\}\).

### Step 2: Apply the AM-GM inequality
We will use the Arithmetic Mean-Geometric Mean (AM-GM) inequality, which states that for any positive numbers \(x\) and \(y\):

\[ \frac{x + y}{2} \geq \sqrt{xy} \]

In our case, we can apply it to \(a\) and \(\frac{1}{a}\):

\[ \frac{a + 1/a}{2} \geq \sqrt{a \cdot 1/a} = 1 \]

This implies:

\[ a + \frac{1}{a} \geq 2 \]

### Step 3: Express the terms
We can rewrite \(a^2 + a + 1\) as:

\[ a^2 + a + 1 = a^2 + a + 1 \geq 3 \sqrt[3]{a^2 \cdot a \cdot 1} = 3 \sqrt[3]{a^3} = 3a \]

Thus, we have:

\[ a^2 + a + 1 \geq 3a \]

### Step 4: Apply this to all terms
Applying this to all five variables, we have:

\[ (a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1)(d^2 + d + 1)(e^2 + e + 1) \geq 3^5 (abcde) \]

### Step 5: Substitute back into the expression
Substituting this back into our expression gives:

\[ \frac{(a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1)(d^2 + d + 1)(e^2 + e + 1)}{abcde} \geq \frac{3^5 (abcde)}{abcde} = 3^5 \]

### Step 6: Calculate the minimum value
Calculating \(3^5\):

\[ 3^5 = 243 \]

Thus, the minimum value of the expression is:

\[ \boxed{243} \]