Infimum And Supremum In Triangles?

In the above triangle, using simple triangle inequality considering triangles, $\text{AGB, BGC, AGC}$ we have,

$\displaystyle \begin{cases} \frac{2}{3}(m_a+m_b) >c \\ \frac{2}{3}(m_a+m_c) >b \\ \frac{2}{3}(m_c+m_b) >a \end{cases}$

Adding we get , $\displaystyle m_a+m_b+m_c > \frac{3}{4}(a+b+c)=12$

Again by construction we claim that $AD=DH$ and since $BD=DC$ we have $ABHC$ a parallelogram which says $BH=AC$

So in $\Delta \text{ABH}$ we have , $2AD<AB+BH=AB+AC$ simillarly , $2BE<BA+BC$ & $2CF<AC+BC$ and adding we have ,

$m_a+m_b+m_c<(a+b+c)=16$ and thus $\displaystyle 12<AA_1+BB_1+CC_1<16$ whch makes the answer $\boxed{16+12=28}$

But to say these are not the max and min of the sum of medians as they would never attain this value whenever the triangle is non-degenerate. So in practical no such Extremum exists which it can attain, this is merely a upper and lower bound found till date which fits perfectly.