For any convex quadrilateral $ABCD$,prove that $AB + BC + CD + DA > AC + BD$.

(N/A) In a triangle,the sum of any two sides is always greater than the third side.
Consider the triangles formed by the diagonals of the quadrilateral $ABCD$:
$1$. In $\triangle ABC$,$AB + BC > AC$ (Equation $1$).
$2$. In $\triangle ADC$,$AD + CD > AC$ (Equation $2$).
$3$. In $\triangle ABD$,$AB + AD > BD$ (Equation $3$).
$4$. In $\triangle BCD$,$BC + CD > BD$ (Equation $4$).
Adding Equations $1$ and $2$ gives: $(AB + BC + AD + CD) > 2AC$.
Adding Equations $3$ and $4$ gives: $(AB + AD + BC + CD) > 2BD$.
Adding these two resulting inequalities: $2(AB + BC + CD + DA) > 2(AC + BD)$.
Dividing by $2$,we get: $AB + BC + CD + DA > AC + BD$.