If the value of the determinant $\left|\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right|$ is positive,  then $(a,b,c>0)$ 

$\begin{array}{l}\Delta =\left|\begin{array}{l}a11\\ 1b1\\ 11c\end{array}\right|=abc-(a+b+c)+2\\ \Delta >0\Rightarrow abc+2>a+b+c\\ abc+2>3{\left(abc\right)}^{\frac{1}{3}}[\because \frac{a+b+c}{3}>{\left(abc\right)}^{\frac{1}{3}}]\end{array}$

$\begin{array}{l}Let{\left(abc\right)}^{\frac{1}{3}}=x\\ x^3+2>3x\Rightarrow x^3-3x+2>0\\ \Rightarrow {(x-1)}^2(x+2)>0\Rightarrow x+2>0\\ \Rightarrow x>-2\Rightarrow {\left(abc\right)}^{\frac{1}{3}}-2\Rightarrow abc>-8\end{array}$