Question

If $\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ba& b^2+1& bc\\ ca& cb& c^2+1\end{array}\right|=k\sqrt{(abc)}$, 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)

$\frac{1}{abc}\left|\begin{array}{ccc}{a}^3+a& a^{2}b& a^{2}c\\ b^{2}a& b^3+b& b^{2}c\\ c^{2}a& c^{2}b& c^3+c\end{array}\right|=\left|\begin{array}{ccc}{a}^2+1& a^2& a^2\\ b^2& b^2+1& b^2\\ c^2& c^2& c^2+1\end{array}\right|=(1+a^2+b^2+$
$c^2)\left|\begin{array}{ccc}1& 1& 1\\ b^2& 1+b^2& b^2\\ c^2& c^2& 1+c^2\end{array}\right|$
$=\left(1+a^2+b^2+c^2\right)=k\sqrt{abc}$
$\text{ Now, }\frac{1+a^2+b^2+c^2}{4}\ge \sqrt{abc}\Rightarrow \frac{1+a^2+b^2+c^2}{\sqrt{abc}}=k\ge 4$