If $\det\left(\begin{smallmatrix}a & 1 & 1\\1 & b & 1\\1 & 1 & c\end{smallmatrix}\right) > 0 $ then prove that $abc> -8$

Question

If the value of the determinant $$\begin{vmatrix} a & 1 & 1\\ 1 & b & 1\\ 1 & 1 & c \end{vmatrix} > 0 $$ then prove that $abc> -8$

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I have calculated its determinant and got

$$abc - (a+b+c) > -2$$

Please tell me how to proceed further to the given answer form.

Answer 1

Another counterexample to this claim is $(a,b,c) = (-1,-1,-k)$ for any $k\ge 8$. Then $$ \begin{vmatrix} a & 1 & 1\\ 1 & b & 1\\ 1 & 1 & c \end{vmatrix} = \begin{vmatrix} -1 & 1 & 1\\ 1 & -1 & 1\\ 1 & 1 & -k \end{vmatrix} = 4 $$ for any choice of $k$, but $abc = -k$ can be made as small as we want.

Answer 2

When $a=b=-3$ and $c=-0.9$, $abc-(a+b+c)=-1.2>-2$ but $abc=-8.1<-8$.