Sum of medians and sum of sides Given a triangle $ABC$, let $D,E,F$ be the midpoints of $BC,CA,AB$ respectively. What are the possible values of $$\frac{AD+BE+CF}{AB+BC+CA}?$$
There is a closed-form formula for the median: $AD=\frac{\sqrt{2AB^2+2AC^2-BC^2}}{2}$, etc. From this it can be shown that $4(AD^2+BE^2+CF^2)=3(AB^2+BC^2+CA^2)$. Can we derive the values of the above ratio from this relation?
 A: The possible range of the ratio $\mathcal{R} \stackrel{def}{=}\frac{AD + BE + CF}{AB + BC + CA}$ is either $\left[\frac34, 1\right]$ of $\left( \frac34, 1\right)$ depends on whether degenerate triangles are allowed or not.


*

*For any pair of points $P,Q$ in the plane. If we fix $P$ and view the distance $PQ$ as a function of $Q$, one is easy to see it is a convex function. This implies
$$AD \le \frac12(AB+AC),\quad BE \le \frac12 (BC+AB)\quad\text{ and }\quad CF \le \frac12 (AC+BC)\tag{*1}$$
Summing these $3$ inequalities leads to
$$AD + BE + CF \le AB + BC + AC \quad\implies\quad \mathcal{R} \le 1$$

*Notice the $3$ medians $AD, BE, CF$ intersect at the centroid $G$ and
$$AD = \frac32 AG,\quad BE = \frac32 BG\quad\text{ and }\quad CF = \frac32 CG$$
Apply triangle inequalities to the three triangles $\triangle GAB$, $\triangle GBC$ and $\triangle GCA$, we have
$$AG + BG \ge AB, \quad BG + CG \ge BC\quad\text{ and }\quad CG + AG \ge AC\tag{*2}$$
Summing these 3 inequalities leads to
$$2(AG+BG+CG) \ge AB + BC +CA \quad\implies\quad
\mathcal{R} = \frac32\frac{AG + BG +CG}{AB +BC +CA} \ge \frac34$$
Combine these, we find $\mathcal{R} \in \left[ \frac34, 1 \right]$ in general.
Notice


*

*$\mathcal{R} = 1$ for a degenerate triangle with sides $1,1,0$

*$\mathcal{R} = \frac34$ for a degenerate triangle with sides $1,1,2$.


When degenerate triangles are allowed, the possible range of $\mathcal{R}$ is indeed $\left[\frac34, 1 \right]$.
When degenerate triangles are not allowed, the inequalities in $(*1)$ and $(*2)$ become strict.
For example,


*

*$AD < \frac12 (AB + AC)$ unless $A$ lies on the line $BC$.

*$AG + BG > AB$ unless $G$ lies on $AB$.


Since one can create non-degenerate triangles with $\mathcal{R}$ as close to $1$ and $\frac34$ as possible, the possible range of $\mathcal{R}$ for non-degenerate triangles is $\left(\frac34, 1\right)$ instead.
