Ans : (A)
Assume that $ a \geq b \geq c$. We must have $b+c > a.$
Also note that $b+c \leq c+a \leq a+b$
$ \large\frac{1}{(b+c)} \geq \large\frac{ 1}{(c+a)} \geq \large\frac{1}{(a+b)}$
To show that $ \large\frac{1}{(b+c)},\large\frac{1}{(c+a)},\large\frac{1}{(a+b)}$ are sides of a triangle,
it is sufficient to show that
$ \large\frac{1}{(c+a)} + \large\frac{1}{(a+b)} > \large\frac{1}{(b+c)}$……………(i)
As $a \geq b \geq c$, we get
$2a \geq a+b\: and\: 2a \geq a+c$
$\large\frac{1}{2a} \leq \large\frac{1}{(a+b)}, \large\frac{1}{2a} \leq \large\frac{1}{(a+c)}$
$\large\frac{1}{(a+b)} + \large\frac{1}{(a+c)} \geq \large\frac{1}{2a} + \large\frac{1}{2a} = \large\frac{1}{a} > \large\frac{1}{(b+c)}$ [From Eq. (i)]
So, it represents a triangle.