For any convex quadrilateral $ABCD$,prove that $AB + BC + CD + DA < 2(AC + BD)$.

(N/A) Let $O$ be the intersection point of the diagonals $AC$ and $BD$.
In $\triangle ABC$,by the triangle inequality,$AB + BC > AC$.
In $\triangle ADC$,by the triangle inequality,$CD + DA > AC$.
Adding these two inequalities,we get: $(AB + BC + CD + DA) > 2AC$.
Similarly,in $\triangle ABD$,$AB + AD > BD$.
In $\triangle BCD$,$BC + CD > BD$.
Adding these two inequalities,we get: $(AB + AD + BC + CD) > 2BD$.
Now,consider the triangles formed by the intersection point $O$:
In $\triangle OAB$,$AB < OA + OB$.
In $\triangle OBC$,$BC < OB + OC$.
In $\triangle OCD$,$CD < OC + OD$.
In $\triangle ODA$,$DA < OD + OA$.
Summing these four inequalities: $AB + BC + CD + DA < 2(OA + OB + OC + OD)$.
Since $OA + OC = AC$ and $OB + OD = BD$,we have $AB + BC + CD + DA < 2(AC + BD)$.