Question

If $a,b,c>0$, the minimum value of $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ is

• A. 1
• B. $\frac{3}{2}$
• C. 2
• D. $\frac{5}{2}$

Answer 1

Answer: (B)

Using $AM\ge GM$
$\frac{\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}{3}\ge \sqrt[3]{\frac{abc}{(a+b)(b+c)(c+a)}}$ ...(i)
Again using $AM\ge GM$
$\frac{a+b}{2}\ge \sqrt{ab},\frac{b+c}{2}\ge \sqrt{bc},\frac{c+a}{2}\ge \sqrt{ca}$
$\Rightarrow \frac{(a+b)(b+c)(c+a)}{abc}\ge 8abc$
$\Rightarrow \sqrt[3]{\frac{abc}{(a+b)(b+c)(c+a)}}\le \frac{1}{2}$
$\therefore$ From Eq. (i)
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$