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.