Question

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

Answer 1

Answer: 81

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