Prove length relationship of median and sides in triangle using triangle inequality AD is the median in the triangle ∆ABC, from the corner A.
Prove that $\frac{AB+AC}{2}$$>AD>$$\frac{AB+AC-BC}{2}$.
I have that
$AB+AC>BC$,
$AB+BC>AC$,
$AC+BC>AB$
as well as 
$AC+AD>CD$
$AD+CD>AC$
$AC+CD>AD$
and
$AB+AD>BD$
$AD+BD>AB$
$AB+BD>AD$
and
$BC=CD+BD$
I come as far as $AB+BC+AC>2AD$ and $AD>AC-BC-BD$ and now I'm really stuck and staring myself blind, going in circles.
Super grateful for any help!

This is just a simple application of the triangle inequality.
Take $A'$ so that $ABCA'$ is a parallelogram. Then $2AD=AA'< AB+BA'=AB+AC$. For the other inequality, we have $AD+DB> AB$ and $AD+DC> AC$. Adding gives $2AD> AB+AC-BC$.
