In triangle $ABC$, the medians $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ concur at the centroid $G$.
(a) Prove that $AD < (AB + AC)/2$.
(b) Let $P=AB+AC+BC$ be the perimeter of $\triangle ABC.$ Prove that $$\frac{3P}{4} < AD + BE + CF < P.$$
I think I should apply the Triangle Inequality in here, but where? Should I connect some points? Make new triangles? I'm stuck, thanks in advance!
Consider the following mystic hexagon:
By applying the triangle inequality to $ABA'$, we get $2\cdot AD<AB+BA' = AB+AC$.
That leads to $AD+BE+CF < AB+AC+BC$.
On the other hand, $B$ is the centroid of $AA'A''$, hence
$$ \frac{3}{2}BA+\frac{3}{2}BA'+\frac{3}{2}BA'' < AA'+AA''+A'A'' = 2(AD+BE+CF).$$