Question

If $\begin{vmatrix}a^2+1 & a b & a c \\ b a & b^2+1 & b c \\ c a & c b & c^2+1\end{vmatrix}=k \sqrt{(a b c)}$, where $a, b, c$ are positive reals then minimum possible value of $k$ is

• A. 1
• B. 4
• C. 16
• D. 64

Answer 1

Answer: (B)

$\frac{1}{a b c}\begin{vmatrix}a^3+a & a^2 b & a^2 c \\ b^2 a & b^3+b & b^2 c \\ c^2 a & c^2 b & c^3+c\end{vmatrix}=\begin{vmatrix}a^2+1 & a^2 & a^2 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1\end{vmatrix}=\left(1+a^2+b^2+\right.$
$\left.c^2\right)\begin{vmatrix}1 & 1 & 1 \\ b^2 & 1+b^2 & b^2 \\ c^2 & c^2 & 1+c^2\end{vmatrix}$
$=\left(1+a^2+b^2+c^2\right)=k \sqrt{a b c} $
$\text { Now, } \frac{1+a^2+b^2+c^2}{4} \geq \sqrt{a b c} \Rightarrow \frac{1+a^2+b^2+c^2}{\sqrt{a b c}}=k \geq 4$