Prove that : $m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$ Let $m_{a},m_{b},m_{c}$ be the lengths of the medians and $a,b,c$ be the lengths of the sides of a given triangle , Prove the inequality : 
$$m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$$
Where : 
$s : \operatorname{Semiperimeter}$
$R : \operatorname{circumradius}$ 
I know the relation : 
$$m_{a}^{2}=\frac{2(b^{2}+c^{2})-a^{2}}{4}$$
But when I multiple together I dont get simple formulas!
So, I need help finding a solution.
Thanks!


Note that the triangles ABD and EDC are similar. Then,
$$\frac{AD}{BD}=\frac{CD}{ED}\implies \frac{m_a}{\frac a2}=\frac{\frac a2}{AE-m_a}
\implies m_a^2 -AE\cdot m_a + \frac {a^2}4=0$$
which, since $AE \le 2R$ and $a=2R\sin A$, leads to
$$m_a =\frac12(AE+\sqrt{AE^2-a^2})\le \frac12\left[2R+\sqrt{(2R)^2-(2R\sin A)^2}\right] =2R\cos^2 \frac A2$$
Likewise, $m_b\le 2R\cos^2 \frac B2$ and $m_c\le 2R\cos^2 \frac C2$. Together, we have
$$\begin{align}
m_a m_bm_c & \le \frac12R^3\left( 4\cos\frac A2\cos \frac B2\cos \frac C2\right)^2  \\
& = \frac12R^3\left( 2 \cos\frac A2 \left(\cos \frac {B+C}2+\cos \frac {B-C}2 \right)\right)^2 \\
& = \frac12R^3\left( 2 \cos\frac A2 \sin\frac A2+2 \sin\frac {B+C}2\cos \frac {B-C}2 \right)^2 \\
& = \frac12R^3\left( \sin A + \sin B + \sin C \right)^2 \\
& = \frac12R^3 \left( \frac a{2R} + \frac b{2R} + \frac c{2R}\right)^2 \\
& = \frac12R \left( \frac{a + b+ c}2 \right)^2 \\
\end{align}$$
Thus,
$$m_a m_bm_c \le \frac12Rs^2$$
