Let point $F$ be the point inside $\triangle{ABC}$ such that $\angle{AFB}=\angle{BFC}=\angle{CFA}=120^\circ$. Also, draw three outwards isosceles triangles from the three sides such that their top angles are all equal to $120^\circ$, as shown:

Consider $\triangle{BFC}$ and $\triangle{BD_1C}$. They share the same base $BC$, and equal corresponding angles ($\angle{BFC}=\angle{BD_1C}$). Because $\triangle{BD_1C}$ is isosceles, it must hold that $$S_{\triangle{BD_1C}} \ge S_{\triangle{BFC}}$$ Similarly, we have $$S_{\triangle{CD_2A}} \ge S_{\triangle{CFA}}\qquad\text{and}\qquad S_{\triangle{AD_3B}} \ge S_{\triangle{AFB}}$$ Adding these three relationships yields \begin{align*} S_{\triangle{BD_1C}} + S_{\triangle{CD_2A}}+S_{\triangle{AD_3C}} & \ge S_{\triangle{BFC}}+S_{\triangle{CFA}}+S_{\triangle{AFB}} \\ \frac{1}{4\sqrt{3}}(a^2 + b^2 +c^2) &\ge S_{\triangle{ABC}}\\ a^2 + b^2 +c^2 &\ge 4\sqrt{3}\cdot S \end{align*}