Question

The minimum value of $\frac{(a^2+3a+1)(b^2+3b+1)(c^2+3c+1)}{abc}$ ,where $a,b,c\in R^{+}$ is

• A. $\frac{{11}^3}{2^3}$
• B. $81$
• C. $25$
• D. $27$

Answer 1

Answer: (B)

$81$

Given expression is,

$(a+\frac{1}{a}+3)(b+\frac{1}{b}+3)(c+\frac{1}{c}+3)$

Now, $AM\ge GM$ for positive quantities.

So, $(a+\frac{1}{a}+3)\ge 3{(3a\times \frac{1}{a})}^{1/3}\left(1\right)(b+\frac{1}{b}+3)\ge 3{(3b\times \frac{1}{b})}^{1/3}\left(2\right)(c+\frac{1}{c}+3)\ge 3{(3c\times \frac{1}{c})}^{1/3}\left(3\right)$

Multiplying $\left(1\right),\left(2\right)$ and $\left(3\right)$

$(a+\frac{1}{a}+3)(b+\frac{1}{b}+3)(c+\frac{1}{c}+3)\ge 3^3\cdot 3(a+\frac{1}{a}+3)(b+\frac{1}{b}+3)(c+\frac{1}{c}+3)\ge 81$