• # question_answer If $a,\ b,\ c$ are the positive integers, then $(a+b)(b+c)(c+a)$is [DCE 2000] A) $<8abc$ B) $>8abc$ C) $=8abc$ D) None of these

Since A.M. > G.M.,  we have $\frac{a+b}{2}>\sqrt{ab},\ \frac{b+c}{2}>\sqrt{bc}$ and $\frac{a+c}{2}>\sqrt{ac}$. Multiplying these inequalities, we get $(a+b)(b+c)(c+a)>8abc$.