Question

The minimum value of $\frac{\left(A^{2} + A + 1\right) \left(B^{2} + B + 1\right) \left(C^{2} + C + 1\right) \left(D^{2} + D + 1\right)}{A B C D}$ , where $A,B,C,D>0$ is

Answer 1

$\frac{\left(A^{2} + A + 1\right) \left(B^{2} + B + 1\right) \left(C^{2} + C + 1\right) \left(D^{2} + D + 1\right)}{A B C D}$
$=\left(A + \frac{1}{A} + 1\right)\left(B + \frac{1}{B} + 1\right)\left(C + \frac{1}{C} + 1\right)\left(D + \frac{1}{D} + 1\right)$
Since $A>0$ ,
So, $\frac{A + \frac{1}{A}}{2}\geq \sqrt{A \times \frac{1}{A}}\Rightarrow A+\frac{1}{A}\geq 2$ (Using AM-GM inequality)
Similarly, $B+\frac{1}{B}\geq 2,C+\frac{1}{C}\geq 2\&D+\frac{1}{D}\geq 2$ .
$\therefore \left(A + \frac{1}{A} + 1\right)\left(B + \frac{1}{B} + 1\right)\left(C + \frac{1}{C} + 1\right)\left(D + \frac{1}{D} + 1\right)$
$\geq \left(2 + 1\right)\times \left(2 + 1\right)\times \left(2 + 1\right)\times \left(2 + 1\right)$
$=3\times 3\times 3\times 3$
$=3^{4}$
$=81$
Hence, $\frac{\left(A^{2} + A + 1\right) \left(B^{2} + B + 1\right) \left(C^{2} + C + 1\right) \left(D^{2} + D + 1\right)}{A B C D}\geq 81$ .
So, the required minimum value is $81$ .