Let $r_b$ and $m_b$ respectively be the exradius of the excircle opposite $B$ and the median drawn from the midpoint of side $CA$ of acute triangles $\triangle ABC$. Prove that $$\large r_a + r_b + r_c \ge m_a + m_b + m_c$$
We have that $$[ABC] = \sqrt{\frac{r_a + r_b + r_c}{2} \cdot \prod_{cyc}\frac{r_a - r_b + r_c}{2}} = \frac{4}{3}\sqrt{\frac{m_a + m_b + m_c}{2} \cdot \prod_{cyc}\frac{m_a - m_b + m_c}{2}}$$
Let $r_a - r_b + r_c = r_b'$, $m_a - m_b + m_c = m_b'$ and so on, we have that $$\sum_{cyc}r_b' \cdot \prod_{cyc}r_b' \ge \frac{16}{9} \cdot \sum_{cyc}m_b' \cdot \prod_{cyc}m_b'$$
In order to prove that $r_a + r_b + r_c \ge m_a + m_b + m_c$, which could be rewritten as $$r_a' + r_b' + r_c' \ge m_a' + m_b' + m_c'$$, we need to prove that $r_a' \cdot r_b' \cdot r_c' \le \dfrac{16}{9} \cdot m_a' \cdot m_b' \cdot m_c'$.
You could do that preferentially... or let $p - a = a'$, $p - b = b'$, $p - c = c'$, we need to prove that $$\sqrt{(a' + b' + c') \cdot (a'b'c')} \cdot \left(\frac{1}{a'} + \frac{1}{b'} + \frac{1}{c'}\right)\ge \sum_{cyc}\sqrt{b'(a' + b' + c') + \frac{(c' - a')^2}{4}}$$
$\left(p = \dfrac{a + b + c}{2}\right)$ in which I don't know what to do next.
Your inequality is true for any triangle!
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and in the standard notation we need to prove that: $$\sum_{cyc}\frac{2S}{b+c-a}\geq\frac{1}{2}\sum_{cyc}\sqrt{2b^2+2c^2-a^2}$$ or $$\sum_{cyc}\frac{2\sqrt{xyz(x+y+z)}}{2x}\geq\frac{1}{2}\sum_{cyc}\sqrt{4x(x+y+z)+(y-z)^2}$$ or $$\frac{2(xy+xz+yz)\sqrt{x+y+z}}{\sqrt{xyz}}\geq\sum_{cyc}\sqrt{4x(x+y+z)+(y-z)^2}$$ or $$\frac{4(xy+xz+yz)^2(x+y+z)}{xyz}\geq\sum_{cyc}(4x(x+y+z)+(y-z)^2)+$$ $$+2\sum_{cyc}\sqrt{(4x(x+y+z)+(y-z)^2)(4y(x+y+z)+(x-z)^2)}$$ or $$\frac{4(xy+xz+yz)^2(x+y+z)}{xyz}-3\sum_{cyc}(4x(x+y+z)+(y-z)^2)+$$ $$+\sum_{cyc}\left(\sqrt{4x(x+y+z)+(y-z)^2}-\sqrt{4y(x+y+z)+(x-z)^2}\right)^2\geq0$$ or $$\frac{4(xy+xz+yz)^2(x+y+z)}{xyz}-18\sum_{cyc}(x^2+xy)+$$ $$+\sum_{cyc}\frac{(x-y)^2(4(x+y+z)-(x+y-2z))^2}{\left(\sqrt{4x(x+y+z)+(y-z)^2}+\sqrt{4y(x+y+z)+(x-z)^2}\right)^2}\geq0,$$ for which it's enough to prove that: $$\frac{4(xy+xz+yz)^2(x+y+z)}{xyz}-18\sum_{cyc}(x^2+xy)+$$ $$+\sum_{cyc}\frac{9(x-y)^2(x+y+2z)^2}{\frac{1}{2}\left(4x(x+y+z)+(y-z)^2+4y(x+y+z)+(x-z)^2\right)}\geq0$$ or $$\frac{2(xy+xz+yz)^2(x+y+z)}{xyz}-9\sum_{cyc}(x^2+xy)+$$ $$+\sum_{cyc}\frac{9(x-y)^2(x+y+2z)^2}{4(x+y)(x+y+z)+(y-z)^2+(x-z)^2}\geq0$$ or $$\sum_{cyc}(2x^3y^2+2x^3z^2-5x^3yz+x^2y^2z)+\sum_{cyc}\tfrac{9xyz(x-y)^2(x+y+2z)^2}{4(x+y)(x+y+z)+(y-z)^2+(x-z)^2}\geq0$$ or $$\sum_{cyc}(2x^3y^2+2x^3z^2-4x^3yz-x^3yz+x^2y^2z)+\sum_{cyc}\tfrac{9xyz(x-y)^2(x+y+2z)^2}{4(x+y)(x+y+z)+(y-z)^2+(x-z)^2}\geq0$$ or $$\sum_{cyc}(x-y)^2\left(2z^3-\frac{1}{2}xyz\right)+\sum_{cyc}\tfrac{9xyz(x-y)^2(x+y+2z)^2}{4(x+y)(x+y+z)+(y-z)^2+(x-z)^2}\geq0$$ or $$\sum_{cyc}(x-y)^2\left(4z^3-xyz+\tfrac{18xyz(x+y+2z)^2}{4(x+y)(x+y+z)+(y-z)^2+(x-z)^2}\right)\geq0,$$ For which it's enough to prove that: $$18(x+y+2z)^2\geq4(x+y)(x+y+z)+(y-z)^2+(x-z)^2$$ or $$13x^2+28xy+13y^2+70(x+y+z)z\geq0$$ and we are done!