Prove that: $ m_a + m_b + m_c \le 4R + r $

Let $M$ be the midpoint of the side $BC$, and let $O$ be the circumcenter.

Then, we can have $$m_a\le AO+OM\le R+R\cos A$$ So, we have $$m_a+m_b+m_c\le 3R+R(\cos A+\cos B+\cos C)$$ Using that $$\cos A+\cos B+\cos C=1+\frac rR$$ we get $$m_a+m_b+m_c\le 4R+r$$

We need to prove that $\sum\limits_{cyc}\left(m_a^2+2m_am_b\right)\leq(4R+r)^2$ and

since by Ptolemy $2m_am_b\leq c^2+\frac{1}{2}ab$, it remains to prove that $$\sum\limits_{cyc}\left(\frac{1}{4}(2b^2+2c^2-a^2)+c^2+\frac{1}{2}ab\right)\leq\left(\frac{abc}{S}+\frac{2S}{a+b+c}\right)^2$$ or $$\sum\limits_{cyc}(7a^2+2ab)\leq4\left(\frac{8abc(a+b+c)+16S^2}{8S(a+b+c)}\right)^2$$ or $$\sum\limits_{cyc}(7a^2+2ab)\leq4\left(\frac{8abc+\prod\limits_{cyc}(a+b-c)}{8S}\right)^2$$ or $$\sum\limits_{cyc}(7a^2+2ab)\leq4\left(\frac{\sum\limits_{cyc}(-a^3+a^2b+a^2c+2abc)}{8S}\right)^2$$ or $$\sum\limits_{cyc}(7a^2+2ab)\leq\frac{\left(\sum\limits_{cyc}(-a^3+abc+a^2b+a^2c+abc)\right)^2}{\sum\limits_{cyc}(2a^2b^2-a^4)}$$ or $$\sum\limits_{cyc}(7a^2+2ab)\leq\frac{\left(\sum\limits_{cyc}(-a^2+ab+ab)\right)^2(a+b+c)}{\prod\limits_{cyc}(a+b-c)}$$ or $$\sum\limits_{cyc}(a^5-a^4b-a^4c+a^3bc)\geq0,$$ which is Schur and it's true even for all positives $a$, $b$ and $c$.

Done!