Question

ABCD is quadrilateral.

Is AB + BC + CD + DA < 2 (AC + BD)?

Answer 1

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.