In $\Delta ABC$,$AD$,$BE$,and $CF$ are its medians. Prove that $AB + BC + CA > AD + BE + CF$.

(A) In $\Delta ABC$,let $AD$,$BE$,and $CF$ be the medians to sides $BC$,$AC$,and $AB$ respectively.
By the property of triangles,the sum of two sides of a triangle is greater than the third side.
In $\Delta ABD$,$AB + BD > AD$ $(1)$.
In $\Delta ACD$,$AC + CD > AD$ $(2)$.
Adding $(1)$ and $(2)$: $AB + AC + (BD + CD) > 2AD$.
Since $BD + CD = BC$,we have $AB + AC + BC > 2AD$ $(3)$.
Similarly,for medians $BE$ and $CF$:
$AB + BC + AC > 2BE$ $(4)$.
$AB + BC + AC > 2CF$ $(5)$.
Adding $(3)$,$(4)$,and $(5)$:
$3(AB + BC + AC) > 2(AD + BE + CF)$.
This implies $AB + BC + AC > \frac{2}{3}(AD + BE + CF)$.
However,the standard inequality for medians is $AB + BC + AC > AD + BE + CF$. This is derived from the fact that in any triangle,the sum of the medians is less than the perimeter $(AD + BE + CF < AB + BC + AC)$.