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 $AM \geq GM$
$\frac{\frac{a}{b +c}+\frac{b}{c +a}+\frac{c}{a +b}}{3} \geq \sqrt[3]{\frac{a b c}{(a +b)(b +c)(c+ a)}}$ ...(i)
Again using $AM \geq GM$
$\frac{a +b}{2} \geq \sqrt{a b}, \frac{b+c}{2} \geq \sqrt{b c}, \frac{c +a}{2} \geq \sqrt{c a}$
$\Rightarrow \frac{(a +b)(b +c)(c +a)}{a b c} \geq 8 a b c$
$\Rightarrow \sqrt[3]{\frac{a b c}{(a +b)(b +c)(c +a)}} \leq \frac{1}{2}$
$\therefore $ From Eq. (i)
$\frac{a}{b+ c}+\frac{b}{c+ a}+\frac{c}{a+ b} \geq \frac{3}{2}$