Let $s$ be the semiperimeter of the triangle, and let $x = s - a$, $y = s - b$ and $z = s - c$. Note that $x$, $y$, $z > 0$ by the triangle inequality. Now,
$$\begin{align*} & a^2 + b^2 + c^2 \ge 4\sqrt{3T} + (a-b)^2 + (b-c)^2 + (c-a)^2 \\ \Leftrightarrow & \quad 2ab + 2bc + 2ca \ge 4\sqrt{3s(s-a)(s-b)(s-c)} + a^2 + b^2 + c^2 \\ \Leftrightarrow & \quad 2 \sum_{\text{cyc}} (y+z)(x+z) \ge 4\sqrt{3(x+y+z)xyz} + (y+z)^2 + (z+x)^2 + (x+y)^2 \\ \Leftrightarrow & \quad 4(xy + yz + zx) \ge 4\sqrt{3(x+y+z)xyz} \\ \Leftrightarrow & \quad (xy + yz + zx)^2 \ge 3((xy)(yz) + (yz)(zx) + (zx)(xy)). \end{align*}$$

This is a well-known inequality (for example, the difference of the left-hand side and the right-hand side is $\frac{1}{2} \sum_{\text{cyc}} (xy - yz)^2$). Equality holds when $xy = yz = zx$, i.e. $a = b = c$.