Explanation

(b)
We have $a,b,c$ are sides of a triangle
$\therefore b + c - a > 0,c + a - b > 0,a + b - c > 0$
Let $x = b + c - a,y = c + a - b,\mathcal{z =}a + b - c$
$\Rightarrow y + \mathcal{z} = 2a,\mathcal{z +}x = 2b,x + y = 2c$
Now, $\frac{a}{b + c - a} + \frac{b}{c + a - b} + \frac{c}{a + b - c}$
$= \frac{y + \mathcal{z}}{2x} + \frac{\mathcal{z +}x}{2y} + \frac{x + y}{2\mathcal{z}}$
$= \frac{1}{2}\left( \frac{x}{y} + \frac{y}{\mathcal{z}} + \frac{\mathcal{z}}{x} + \frac{x}{\mathcal{z}} + \frac{y}{x} + \frac{\mathcal{z}}{y} \right)$
$\geq \frac{6}{2}\left( \frac{y}{x} \bullet \frac{x}{y} \bullet \frac{y}{\mathcal{z}} \bullet \frac{\mathcal{z}}{y} \bullet \frac{\mathcal{z}}{x} \bullet \frac{x}{\mathcal{z}} \right)\ \ \ (\because AM \geq GM)$
$= 3$