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

It is given in the question that,

A quadrilateral ABCD, in which we have to show that:

Whether AB + BC + CD + DA < 2 (AC + BD) or not

We know that,

In a triangle, the sum of the length of either two sides of the triangle is always greater than the third side

In ∆OAB, we have

OA + OB > AB (i)

In ∆OBC, we have

OB + OC > BC (ii)

In ∆OCD, we have

OC + OD > CD (iii)

In ∆ODA, we have

OD + OA > DA (iv)

Adding equations (i), (ii), (iii) and (iv), we get:

OA + OB + OB + OC + OC + OD + OD + OA > AB + BC + CD + DA

2 OA + 2 OB + 2 OC + 2 OD > AB + BC + CD + DA

2 OA + 2 OC + 2 OB + 2 OD > AB + BC + CD + DA

2 (0A + OC) + 2 (OB + OD) > AB + BC + CD + DA

2 (AC) + 2 (BD) > AB + BC + CD + DA

2 (AC + BD) >AB + BC + CD + DA

Hence, the expression given in the question is true