See also http://www.mathlinks.ro/Forum/viewtopic.php?t=199673 (with not quite the nicest solution). dg

Let $ a \leq b \leq c$. Hence, we need to show that \[ b^2 - a^2 - c^2 + 4ac \geq 4\sqrt {3}S\,.\] However, since $ 2ac\cos(B) = a^2 + c^2 - b^2$ and $ 2S = ac\sin(B)$, the inequality is equivalent to \[ ac\big(2 - \cos(B)\big) \geq \sqrt {3}ac\sin(B)\,.\] Thus, it suffices to show \[ \cos(B) + \sqrt {3}\sin(B) \leq 2\,,\] but this can be done using Cauchy-Schwarz Ineq, or expanding $ \cos\left(B - 60^\circ\right)$. In addition, the equality holds iff the second largest angle is equal to $ 60^\circ$, which, in this case, is $ \angle{B}$.