Prove that: $ m_a + m_b + m_c \le 4R + r $ 
Let $m_a, m_b, m_c - $ are median of triangle $ABC$.
  Prove that:
  $$ m_a + m_b + m_c \le 4R + r $$

I used 
$$m_a^2=\frac{2b^2+2c^2-a^2}{4}$$
$$R=\frac{abc}{4R}$$
$$r=\frac{2S}{a+b+c}$$
and $R \ge 2r$
 A: 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
$$m_a+m_b+m_c\le 4R+r$$
A: 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!
