Question

If $a>0,b>0,c>0$, then $(a+b)(b+c)(c+a)$ is greater than

• A. $2(a+b+c)$
• B. $3(a+b+c)$
• C. $6abc$
• D. $8abc$

Answer 1

Answer: (D)

Since A.M. > G.M.

$\therefore \frac{a+b}{2}>\sqrt{ab};\frac{b+c}{2}>\sqrt{bc};\frac{c+a}{2}>\sqrt{ca}$

$\therefore \frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8}>\sqrt{ab}\sqrt{bc}\sqrt{ca}=abc$

$\therefore \left(a+b\right)\left(b+c\right)\left(c+a\right)>8abc$