Show that in a quadrilateral $ABCD$,$AB + BC + CD + DA > AC + BD$.

(N/A) Given: $A$ quadrilateral $ABCD$.
To prove: $AB + BC + CD + DA > AC + BD$.
Proof: In $\triangle ABC$,we have $AB + BC > AC$ $\ldots(1)$ [Since the sum of the lengths of any two sides of a triangle must be greater than the third side].
In $\triangle BCD$,we have $BC + CD > BD$ $\ldots(2)$ [Same reason].
In $\triangle CDA$,we have $CD + DA > AC$ $\ldots(3)$ [Same reason].
In $\triangle DAB$,we have $DA + AB > BD$ $\ldots(4)$ [Same reason].
Adding $(1), (2), (3)$ and $(4)$,we get:
$(AB + BC) + (BC + CD) + (CD + DA) + (DA + AB) > AC + BD + AC + BD$
$2AB + 2BC + 2CD + 2DA > 2AC + 2BD$
$2(AB + BC + CD + DA) > 2(AC + BD)$
$AB + BC + CD + DA > AC + BD$.
Hence,proved.