Q72 / 125 · BITSAT 2020 · Mathematics
If $a>0, b>0, c>0$ and $a, b, c$ are distinct, then $(a+$b) $(b+c)(c+a)$ is greater than
- A$2(a+b+c)$
- B$3(a+b+c)$
- C$6 \mathrm{abc}$
- Correct answerD$8 \mathrm{abc}$
Solution
$\mathrm{AM}>\mathrm{GM}$
$\Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2}>\sqrt{\mathrm{ab}}, \frac{\mathrm{b}+\mathrm{c}}{2}>\sqrt{\mathrm{bc}}, \frac{\mathrm{c}+\mathrm{a}}{2}>\sqrt{\mathrm{ca}}$
So, $(a+b)(b+c)(c+a)>8 a b c$
Hence, the correct answer is option: 'D'.