In a triangle, the sum of the lengths of either two sides is always greater than the third side.
Considering ΔOAB,
OA+OB>AB (i)
In ΔOBC,
OB+OC>BC (ii)
In ΔOCD,
OC+OD>CD (iii)
In ΔODA,
OD+OA>DA (iv)
Adding equations (i), (ii), (iii), and (iv), we obtain
OA+OB+OB+OC+OC+OD+OD+OA>AB+BC+CD+DA
⇒ 2OA+2OB+2OC+2OD>AB+BC+CD+DA
⇒ 2OA+2OC+2OB+2OD>AB+BC+CD+DA
⇒ 2(OA+OC)+2(OB+OD)>AB+BC+CD+DA
⇒ 2(AC)+2(BD)>AB+BC+CD+DA
⇒ 2(AC+BD)>AB+BC+CD+DA
Yes, the given expression is true.