<s>[INST] If $\det\left(\begin{smallmatrix}a & 1 & 1\\1 & b & 1\\1 & 1 & c\end{smallmatrix}\right) > 0 $ then prove that $abc> -8$ 
If the value of the determinant $$\begin{vmatrix}
a & 1 & 1\\
1 & b & 1\\
1 & 1 & c
\end{vmatrix} > 0 $$ then prove that $abc> -8$


I have calculated its determinant and got
$$abc - (a+b+c) > -2$$
Please tell me how to proceed further to the given answer form.
 [/INST] 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.
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