Question

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

• A. $\frac{{11}^3}{2^3}$
• B. $125$
• C. $25$
• D. $27$

Answer 1

The correct option is B $125$

A fraction has minimum value when the numerator has minimum value.

As $a,b$ & $c$ $\in$ $R$

$\therefore$ each of $a,b$ & $c$ will assume the value $=1$.

$\therefore$ putting $(a,b,c)=(1,1,1)$

In the given expression,

$\frac{(a^2+3a+1)(b^2+3b+1)(c^2+3c+1)}{abc}$

$=$ $\frac{(1^2+3\times 1^2+1)\times (1^2+3\times 1^2+1)\times (1^2+3\times 1^2+1)}{1\times 1\times 1}=\frac{5\times 5\times 5}{1}=125$

Ans- Option B