Question

The minimum value of $\frac{(A^2+A+1)(B^2+B+1)(C^2+C+1)(D^2+D+1)}{BACD}$

• A. $3^4$
• B. $3^{-4}$
• C. $2^4$
• D. $2^{-4}$

Answer 1

Answer: (A)

$3^4$

$\frac{(A^2+A+1)(B^2+B+1)(C^2+C+1)(D^2+D+1)}{BACD}$

$=\frac{(A^2+A+1)}{A}\frac{(B^2+B+1)}{B}\frac{(C^2+C+1)}{C}\frac{(D^2+D+1)}{D}$

$=(A+\frac{1}{A}+1)(B+\frac{1}{B}+1)(C+\frac{1}{C}+1)(D+\frac{1}{D}+1)$

Use AM & GM

$\frac{A+\frac{1}{A}}{2}\ge 1$

$A+\frac{1}{A}\ge 2$

(I) $(A+\frac{1}{A}+1)\ge 3$

(II) similarly $(B+\frac{1}{B}+1)\ge 3$

(III) $(C+\frac{1}{C}+1)\ge 3$

(IV) $(D+\frac{1}{D}+1)\ge 3$

Multiply them

$(A+\frac{1}{A}+1)(B+\frac{1}{B}+1)(C+\frac{1}{C}+1)(D+\frac{1}{D}+1)\ge 3\times 3\times 3\times 3$

$(A+\frac{1}{A}+1)(B+\frac{1}{B}+1)(C+\frac{1}{C}+1)(D+\frac{1}{D}+1)\ge 3^4$

$\therefore \frac{(A^2+A+1)(B^2+B+1)(C^2+C+1)(D^2+D+1)}{BACD}\ge 3^4$

So min value is $3^4$