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