Let $a$, $b$ and $c$ be sides-lengths, $m_a$, $m_b$ and $m_c$ be medians-lengths
and $\alpha$, $\beta$ and $\gamma$ be measured-angles of the triangle. Prove that: $$m_a\cos\frac{\alpha}{2}+m_b\cos\frac{\beta}{2}+m_c\cos\frac{\gamma}{2}\geq\frac{3}{4}(a+b+c)$$
I tried Rearrangement, SOS and more.
I think, we can kill this inequality by Holder, but it's a very ugly way.
The equality "occurs" also for $a=b=1$ and $c\rightarrow2^-$, which adds problems.
I can give a hint. You can prove that for any triangle $ABC$ and point $P$ the following inequality holds: $$\sum PA.\cos(A/2) \geq p$$ To show this use Bottema's inequality for two triangles (e.g. see here). Use that fact that the cosines of the half-angles can be sides of a triangle and plug them into Bottema's inequality. The RHS of the result will be not less than $p$ but buckle up for some tedious transformations.
The equality occurs in the case you mention because at one point there is a term $\sum \cos(A)$ in front of $p^2$, which is strictly greater than $1$ unless the triangle is degenerate.
In the special case when P is circumcentert this is easily proven as follows. Start from $R \geq 2r$ and use $abc=4Rrp$. Hence $2pR^2 \geq 4Rpr$ which is equivalent to $R^2(a+b+c) \geq abc $ and using the law of sines we obtain $$\sin(A)+\sin(B)+\sin(C) \geq 4\sin(A)\sin(B)\sin(C)$$ Now using the transform $A$ into $(\pi - A)/2$, etc. to get $$\sum \cos(A/2) \geq 4\prod \cos(A/2)$$. The last product of cosines is equal to $p/4R$ hence this special case follows: $$\sum \cos(A/2) \geq p/R $$