It is easy to see (by sine version of area for example), that $[DAI]=[HCG]=[EBF]=[ABC]$. So, if you let the sidelengths of $ABC$ be $a,b,c$, we want to find max $k$ such that $a^2+b^2+c^2+4[ABC]\ge k[ABC]$ So we want to find max $k-4=q$ such that $a^2+b^2+c^2\ge q \sqrt{s(s-a)(s-b)(s-c)}$ (Herons formula) Expand fully, this simplifies to $(\frac{16}{q^2}+1)(a^4+b^4+c^4)\ge (2-\frac{32}{q^2})(a^2b^2+a^2c^2+b^2c^2)$ This is obvious when $q=4\sqrt{3}$ by AMGM. $k=4\sqrt{3}+4$ can be achieved when $ABC$ is equilateral. Thus, the answer is $k=4\sqrt{3}+4$ This is quite nice.