If the value of the determinant $\begin{vmatrix}a&1&1\\1 &b& 1\\1&1&c\end{vmatrix}$ is positive, then
Answer & explanation
Correct answer: option 2
We have,
$Δ=\begin{vmatrix}a&1&1\\1 &b& 1\\1&1&c\end{vmatrix}=abc - (a+b+c) +2$
$∴Δ>0$
$⇒abc +2> a+b+c$
$⇒abc+2>3(abc)^{1/3}$ $\left[∵A.M.>G.M.⇒\frac{a+b+c}{3}>(abc)^{1/3}\right]$
$x^3+2> 3x$, where $x =(abc)^{1/3}$
$⇒x^3-3x+2>0$
$⇒(x-1)^2 (x+2) > 0$
$⇒ x+2>0⇒ x>-2⇒ (abc)^{1/3} >-2 ⇒ abc >-8$