Question

The least value of the product $xyz,(x,y,z\ge 0$ and $x,y,z\ne 1)$ for which the determinant $\left|\begin{array}{ccc}x& 1& 1\\ 1& y& 1\\ 1& 1& z\end{array}\right|$ is non-negative is:

• A. $8$
• B. $0$
• C. $2\sqrt{2}$
• D. $16\sqrt{2}$

Answer 1

The correct option is B $0$

$\left|\begin{array}{ccc}x& 1& 1\\ 1& y& 1\\ 1& 1& z\end{array}\right|\ge 0$
$\Rightarrow xyz-x+1-z+1-y\ge 0$
$\Rightarrow xyz-(x+y+z)+2\ge 0$
$\Rightarrow xyz+2\ge (x+y+z)$

As $\text{A.M.}\ge \text{G.M.}$
$\Rightarrow \frac{x+y+z}{3}\ge \sqrt[3]{3xyz}$

$\therefore xyz+2\ge (x+y+z)\ge 3\cdot \sqrt[3]{3xyz}$
Let $z=\sqrt[3]{3xyz}$

$\therefore z^3-3z+2\ge 0$
$\Rightarrow (z-1{)}^2(z+2)\ge 0$
$\Rightarrow z+2\ge 0(\because (z-1{)}^2\ge 0)$
$\Rightarrow \sqrt[3]{3xyz}\ge -2$
$\Rightarrow xyz\ge -8$
Since, $x,y,z\ge 0$
Hence, minimum value of $xyz=0$