<s>[INST] Triangle inequality with the exradii $r_{a}$, $r_{b}$, $r_{c}$, the medians $m_{a}$, $m_{b}$, $m_{c}$ Given a triangle with the exradii $r_{a}$, $r_{b}$, $r_{c}$, the medians $m_{a}$, $m_{b}$, $m_{c}$. Show that
$$r_{a}^{2}+ r_{b}^{2}+ r_{c}^{2}\geq  3\sqrt{3}. S+ \left ( m_{a}- m_{b} \right )^{2}+ \left ( m_{b}- m_{c} \right )^{2}+ \left ( m_{c}- m_{a} \right )^{2}$$
I can only show the weaker inequality 
$$r_{a}^{2}+ r_{b}^{2}+ r_{c}^{2}\geq  3\sqrt{3}. S$$
I need to the help. Thanks!
 [/INST] There is another proof bases on $R, r, s$ relationship as follows:
Let $R, r, s$ be the circumcircle radius, incircle radius, and semi-perimeter of the triangle. Then  $a, b, c, r_a, r_b, r_c$ satisfy the following relationship:
$$ R\ge 2r$$
$$ r_a+r_b+r_c=4R+r $$
$$r_ar_b+r_br_c+r_cr_a=s^2 $$
$$m_a^2+m_b^2+m_c^2=\dfrac{3}{4}(a^2+b^2+c^2) $$
$$ a^2+b^2+c^2=2(s^2-4Rr-r^2)$$
$$ s^2\le 4R^2+4Rr+3r^2\ \text{(Gerretsen inequality)}$$
We have: $R\ge 2r$ then: $4R^2\ge 16r^2$. Thus:
$$32R^2+28Rr+5r^2\ge 28R^2+28Rr+21r^2$$
By Gerretsen inequality we have: $32R^2+28Rr+5r^2\ge 7s^2$.
Hence: $32R^2+16Rr+2r^2+12Rr+3r^2\ge7s^2$ or equivalently:
$$2(4R+r)^2+12Rr+3r^2\ge 7s^2$$
$$\Leftrightarrow 2(4R+r)^2-4s^2\ge 3s^2-12Rr-3r^2$$
$$\Leftrightarrow 2\Big((4R+r)^2-2s^2\Big)\ge 3(s^2-4Rr-r^2)$$
$$\Leftrightarrow (4R+r)^2-2s^2\ge \dfrac{3}{2}(s^2-4Rr-r^2)$$
$$ \Leftrightarrow r_a^2+r_b^2+r_c^2\ge 2\cdot\dfrac{3}{4}(a^2+b^2+c^2)=m_a^2+m_b^2+m_c^2$$
Now, note that $m_a, m_b, m_c$ are also the sidelength of a triangle, with area equals to $\dfrac{3}{4}S$, then by the well-known Finsler-Hadwiger inequality we have:
$$ m_a^2+m_b^2+m_c^2 \ge 3\sqrt{3}S+(m_a-m_b)^2+(m_b-m_c)^2+(m_c-m_a)^2 $$
Combine the two results above we obtain:  $$ r_a^2+r_b^2+r_c^2 \ge 3\sqrt{3}S+(m_a-m_b)^2+(m_b-m_c)^2+(m_c-m_a)^2 $$
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