Given a triangle ABC and D is the midpoint of BC. Prove that
(a) AD < (AB+AC)/2
(b) AD > (AB+AC-BC)/2
Using vectors this is rather easy.
$\displaystyle \overrightarrow {DA} = \overrightarrow {AB} + \overrightarrow {BD} \;\& \,\overrightarrow {DA} = \overrightarrow {AC} + \overrightarrow {CD} $
Note that $\displaystyle \overrightarrow {BD}+\overrightarrow {CD}=0$.
Use the triangle inequality.