Prove length relationship of median and sides in triangle using triangle inequality

Question

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!

Answer 1

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 $2AD> AB+AC-BC$.