Let a, b, c be distinct real numbers and D be the determinant given by $D=\begin{vmatrix}a&1&1\\1&b&1\\1&1&c\end{vmatrix}$
Statement-1: If $D >0$, then $abc >-8$
Statement-2: A.M.>G.M.
Answer & explanation
Correct answer: option 1
We have,
$D=\begin{vmatrix}a&1&1\\1&b&1\\1&1&c\end{vmatrix}=abc-(a+b+c) + 2$
$∴D >0$
$⇒abc - (a+b+c)+2>0$
$⇒abc +2> a+b+c$ ...(i)
But, A.M. >G.M.
$⇒\frac{a+b+c}{3}>(abc)^{1/3}$
$⇒a+b+c>(abc)^{1/3}$ ...(ii)
From (i) and (ii), we get
$abc+2> 3 (abc)^{1/3}$
$⇒ x^3+2>3x$, where $x = (abc)^{1/3}$
$⇒ x^3-3x+2>0 (x-1)^2 (x+2) > 0$
$⇒ x+2>0⇒ (abc)^{1/3}+2>0$
$⇒ (abc)^{1/3} >-2⇒ abc >-8$