Explanation

(b)
$\because AM > GM$
$\frac{(a + b - c) + (b + c - a)}{2} > \sqrt{(a + b - c)(b + c - a)}$
$\Rightarrow b > \sqrt{(a + b - c)(b + c - a)}\ \ \$…(i)
Similarly, $\frac{(b + c - a) + (c + a - b)}{2} > \sqrt{(b + c - a)(c + a - b)}$
$\Rightarrow c > \sqrt{(b + c - a)(c + a - b)}\ \ \ \$…(ii)
and $\frac{(c + a - b) + (a + b - c)}{2} > \sqrt{(c + a - b)(a + b - c)}$
$\Rightarrow a > \sqrt{(c + a - b)(a + b - c)}$ …(iii)
On multiplying relations (i), (ii) and (iii), we get
$abc > (a + b - c)(b + c - a)(c + a - b)$
$\Rightarrow (a + b - c)(b + c - a)(c + a - b) - abc < 0$