Q45 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics
If $\left|\begin{array}{lll}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right| \gt 0$, then $a b c \gt$
- A1
- Correct answerB-8
- C8
- D3
Solution
Since, $\left|\begin{array}{ccc}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right| \gt 0$
$a(b c-1)-(c-1)+(1-b) \gt 0 \Rightarrow 2+a b c \gt a+b+c$ ....(i)
Since, $A . M . \geq G . M . \Rightarrow a+b+c \geq 3(a b c)^{1 / 3}$ ....(ii)
$\begin{aligned} & \text {By (i) and (ii), we get } 2+a b c \gt 3(a b c)^{1 / 3} \\ & \Rightarrow 2+x^3 \gt 3 x ;\left[\text {Let, }(a b c)^{1 / 3}=x\right] \\ & \Rightarrow(x-1)\left(x^2+x-2\right) \gt 0 \Rightarrow(x-1)^2(x+2) \gt 0 \\ & \Rightarrow x \gt -2 \Rightarrow(a b c)^{1 / 3} \gt -2 \Rightarrow(a b c) \gt -8\end{aligned}$
Hence, the correct answer is option: 'B'.