Explanation

(d)
Using, AM > GM
$\therefore\ \ \frac{a + b + c}{3} > \sqrt[3]{abc}$
$\Rightarrow a + b + c > 3$ ...(i)
[$\therefore abc = 1$ given]
Also, GM > HM
$\sqrt[3]{abc} > \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$
$\Rightarrow \ \ \ \ (1)^{1/3} > \frac{3abc}{bc + ac + ab}$
$\Rightarrow \ \ ab + bc + ac > 3$ ...(ii)
$\therefore$ From Eqs. (i) and (ii), we get
$a + b + c + ab + bc + ac > 6$