ABCD is quadrilateral.
Is AB + BC + CD + DA < 2 (AC + BD)?
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.