Question

The least value of the product $xyz$ 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. $-2\sqrt{2}$
• B. $-16\sqrt{2}$
• C. $-8$
• D. $-1$

Answer 1

Answer: (C)

Where $\left|\begin{array}{ccc}x& 1& 1\\ 1& y& 1\\ 1& 1& z\end{array}\right|\ge 0$
$\Rightarrow x(yz-1)-1(z-1)+(1-y)=0$
$\Rightarrow xyz-x-y-z+2\ge 0$
$\Rightarrow xyz\ge x+y+z-2$...(1)
For $x,y,z,\frac{x+y+z}{3}\ge \sqrt[3]{xyz}$
$\Rightarrow x+y+z\ge 3\cdot \sqrt[3]{xyz}$...(2)
From (1) and (2)
$xyz\ge 3\cdot \sqrt[3]{xyz}-2$
$\Rightarrow t^3\ge 3t-2$ where $t=\sqrt[3]{xyz}$
$\Rightarrow t^3-3t+2\ge 0$
$\Rightarrow (t-1)\left(t^2+t-2\right)\ge 0$
$\Rightarrow (t-1)(t+2)(t-1)\ge 0$
$\Rightarrow (t-1{)}^2(t+2)\ge 0$
$\Rightarrow t+2\ge 0$
$\Rightarrow t\ge -2$
$\Rightarrow \sqrt[3]{xyz}\ge -2$
$\Rightarrow xyz\ge -8$