In $\Delta ABC$,$\overline{AD}$,$\overline{BE}$ and $\overline{CF}$ are medians. Prove that $3(AB^2 + BC^2 + AC^2) = 4(AD^2 + BE^2 + CF^2)$.

(N/A) In $\Delta ABC$,$\overline{AD}$,$\overline{BE}$ and $\overline{CF}$ are medians. Therefore,$D$,$E$,and $F$ are the midpoints of $\overline{BC}$,$\overline{AC}$,and $\overline{AB}$ respectively.
By Apollonius' theorem in $\Delta ABC$ with median $\overline{AD}$:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
Since $BD = \frac{1}{2}BC$,we have:
$AB^2 + AC^2 = 2AD^2 + 2(\frac{BC}{2})^2 = 2AD^2 + \frac{BC^2}{2}$
$2AD^2 = AB^2 + AC^2 - \frac{BC^2}{2} \quad \dots(1)$
Similarly,for medians $\overline{BE}$ and $\overline{CF}$:
$2BE^2 = AB^2 + BC^2 - \frac{AC^2}{2} \quad \dots(2)$
$2CF^2 = AC^2 + BC^2 - \frac{AB^2}{2} \quad \dots(3)$
Adding equations $(1)$,$(2)$,and $(3)$:
$2(AD^2 + BE^2 + CF^2) = (AB^2 + AB^2 - \frac{AB^2}{2}) + (BC^2 + BC^2 - \frac{BC^2}{2}) + (AC^2 + AC^2 - \frac{AC^2}{2})$
$2(AD^2 + BE^2 + CF^2) = \frac{3}{2}AB^2 + \frac{3}{2}BC^2 + \frac{3}{2}AC^2$
$2(AD^2 + BE^2 + CF^2) = \frac{3}{2}(AB^2 + BC^2 + AC^2)$
Multiplying both sides by $2$:
$4(AD^2 + BE^2 + CF^2) = 3(AB^2 + BC^2 + AC^2)$