The minimum value of `((A^(2) +A+1) (B^(2) +B+1) (C^(2) +C+1 )(D^(2) +D+1 ))/(ABCD)` where `A,B,C,D gt 0` is :

Options

A. `(1)/(3 ^(4))`
B. `(1 )/(2^(4))`
C. `2 ^(4)`
D. `3 ^(4)`

Answer

To find the minimum value of the expression

\[ \frac{(A^2 + A + 1)(B^2 + B + 1)(C^2 + C + 1)(D^2 + D + 1)}{ABCD} \]

where \(A, B, C, D > 0\), we can follow these steps:

### Step 1: Rewrite the expression
We can rewrite the expression as:

\[ \frac{(A^2 + A + 1)}{A} \cdot \frac{(B^2 + B + 1)}{B} \cdot \frac{(C^2 + C + 1)}{C} \cdot \frac{(D^2 + D + 1)}{D} \]

This simplifies to:

\[ \left( \frac{A^2 + A + 1}{A} \right) \left( \frac{B^2 + B + 1}{B} \right) \left( \frac{C^2 + C + 1}{C} \right) \left( \frac{D^2 + D + 1}{D} \right) \]

### Step 2: Simplify each term
Now, we can simplify each term:

\[ \frac{A^2 + A + 1}{A} = A + 1 + \frac{1}{A} \]

Thus, the expression becomes:

\[ \left( A + 1 + \frac{1}{A} \right) \left( B + 1 + \frac{1}{B} \right) \left( C + 1 + \frac{1}{C} \right) \left( D + 1 + \frac{1}{D} \right) \]

### Step 3: Apply AM-GM inequality
Using the AM-GM inequality, we know that for any positive \(x\):

\[ x + 1 + \frac{1}{x} \geq 3 \]

This is because:

\[ \frac{x + 1 + \frac{1}{x}}{3} \geq \sqrt[3]{x \cdot 1 \cdot \frac{1}{x}} = 1 \]

So, we have:

\[ x + 1 + \frac{1}{x} \geq 3 \]

### Step 4: Apply AM-GM to all variables
Applying this to \(A\), \(B\), \(C\), and \(D\):

\[ A + 1 + \frac{1}{A} \geq 3 \]
\[ B + 1 + \frac{1}{B} \geq 3 \]
\[ C + 1 + \frac{1}{C} \geq 3 \]
\[ D + 1 + \frac{1}{D} \geq 3 \]

### Step 5: Combine the inequalities
Multiplying these inequalities together gives:

\[ \left( A + 1 + \frac{1}{A} \right) \left( B + 1 + \frac{1}{B} \right) \left( C + 1 + \frac{1}{C} \right) \left( D + 1 + \frac{1}{D} \right) \geq 3^4 = 81 \]

### Step 6: Conclusion
Thus, the minimum value of the original expression is:

\[ \frac{81}{1} = 81 \]

### Final Answer
The minimum value of

\[ \frac{(A^2 + A + 1)(B^2 + B + 1)(C^2 + C + 1)(D^2 + D + 1)}{ABCD} \]

is \(81\).

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