Consider the ∆ ABC
AB + BC > CA — 1
( Sum of the length of any 2 sides of a triangle is greater than the 3rd side.)
In ∆ ACD .
AD + CD > AC — 2
( Sum of the length of any 2 sides of a triangle is greater than the 3rd side.)
Adding 1 and 2
AB + BC + CD + DA > 2AC — 3
In ∆ ABD
AB + DA > BD — 4
( Sum of the length of any 2 sides of a triangle is greater than the 3rd side.)
In ∆ BCD
BC + CD > BD — 5
( Sum of the length of any 2 sides of a triangle is greater than the 3rd side.)
Adding 4 and 5
AB + DA + BC + CD > 2BD — 6
Adding 3 and 6
2(AB + BC + CD + DA) > 2(AC + BD)
∴ AB + BC + CD + DA > AC + BD (yes, it is proved)
