In a triangle, the sum of the lengths of either two sides is always greater than the third side.
Considering ΔABC,
AB+BC>CA (i)
In ΔBCD,
BC+CD>DB (ii)
In ΔCDA,
CD+DA>AC (iii)
In ΔDAB,
DA+AB>DB (iv)
Adding equations (i), (ii), (iii), and (iv), we obtain
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)
Yes, the given expression is true.