Explanation

(a)
We have, $4a^{2} + 9b^{2} + 16c^{2} - 6ab - 12bc - 8ac = 0$
$\Rightarrow 8a^{2} + 18b^{2} + 32c^{2} - 12ab - 24bc - 16ac = 0$
$\Rightarrow 4a^{2} + 9b^{2} - 12ab + 9b^{2} + 16c^{2} - 24bc + 16c^{2} + 4a^{2} - 16ac = 0$
$\Rightarrow (2a - 3b)^{2} + (3b - 4c)^{2} + (4c - 2a)^{2} = 0$
$\Rightarrow 2a = 3b = 4c = k$
$\Rightarrow a = \frac{k}{2},\ b = \frac{k}{3},\ c = \frac{k}{4}$
$\Rightarrow a,\ b,\ c$ are in HP
GM $\geq$ HM
$\therefore\ \sqrt{ac}\ \geq b$