In ΔABC
AB + BC > AC …(i) (the sum of any two sides of a Δ is greater than third side)
Also in ΔADC
AD + DC > AC …(ii) (reason as above)
Similarly in ΔABD and ΔBCD
AB + AD > BD …(iii)
BC + CD >BD …(iv)
Adding (i), (ii), (iii) and (iv), we get
2AB + 2BC + 2AD + 2CD > 2AC + 2BD
⇒ AB + BC + CD + DA > AC + BD
Hence proved.