inequality between median length and perimeter Is there an inequality between the sum of median lengths and the perimeter? If there is, can you specify a proof as well? I need to use this to solve a question.
I tried using Apollonius theorem. 
$$m_a=\sqrt{\frac{2b^2+2c^2-a^2}{4}}$$

Yes, there is : 
$$\frac 34(\text{the perimeter})\lt \text{(the sum of median lengths)}\lt \text{(the perimeter)}$$
Proof : 
Let $G$ be the centroid of $\triangle{ABC}$, and let $X,Y,Z$ be the midpoint of the side $BC,CA,AB$ respectively.
First of all,
$$GY+GZ\gt YZ,\quad GZ+GX\gt ZX,\quad GX+GY\gt XY$$
and so
$$2(GX+GY+GZ)\gt XY+YZ+ZX=\frac 12(AB+BC+CA),$$
i.e.
$$AX+BY+CZ\gt \frac 34(AB+BC+CA)\tag1$$
Also, since we have
$$AX\lt AZ+ZX=AZ+AY$$
$$BY\lt BX+XY=BX+BZ$$
$$CZ\lt CY+YZ=CY+CX$$
we can have
$$AX+BY+CZ\lt (BX+CX)+(CY+AY)+(AZ+BZ)=AB+BC+CA\tag2$$
From $(1)(2)$,
$$\frac 34(AB+BC+CA)\lt AX+BY+CZ\lt AB+BC+CA,$$
i.e.
$$\frac 34(\text{the perimeter})\lt \text{(the sum of median lengths)}\lt \text{(the perimeter)}$$
