Explanation

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