We have $r_{bc}=\frac{r}{\cos^2{\frac{\alpha}{2}}}$ and so on where $r, \alpha$ are the inradius and the angle of $\triangle ABC$. Then we need to prove that $$\sum_{cyc}m_a^2\cos^2\frac{\alpha}{2}\ge \frac{27\sqrt3}{8}r\sqrt[3]{abc}.$$Note that $$\sum_{cyc}m_a^2\cos^2\frac{\alpha}{2}=\sum_{cyc}\frac{2b^2+2c^2-a^2}{4}\cdot \frac{b^2+c^2-a^2+2bc}{4bc}\ge \sum_{cyc}\frac{(b^2+c^2+2bc-a^2)^2}{16bc}=$$$$=\sum_{cyc}\frac{(a+b+c)^2(b+c-a)^2}{16bc}=\frac{(a+b+c)^2}{16abc}\left[a^3+b^3+c^3+6abc-ab(a+b)-bc(b+c)-ca(c+a)\right]\ge$$$$\ge \frac{(a+b+c)^2}{16abc}\cdot 3abc\ge \frac{9}{16}(a+b+c)\sqrt[3]{abc}\ge \frac{27\sqrt3}{8}r\sqrt[3]{abc}$$where the last inequality $a+b+c\ge 6\sqrt3 r$ is true due to $$r^2=\frac{(p-a)(p-b)(p-c)}{p}\le \frac{1}{p}\cdot \left(\frac{p-a+p-b+p-c}{3}\right)^3=\frac{p^2}{27}$$where $2p=a+b+c$.