ABCD is a quadrilateral.

Is AB + BC + CD + DA > AC + BD?

It is known that the sum of the lengths of any two sides of a triangle is greater than the length of the third side

So, in ΔABC,

AB + BC > CA … [i]

Now in ΔBCD

BC + CD > DB … [ii]

and in ΔCDA

CD + DA > AC … [iii]

and in ΔDAB

DA + AB > DB … [iv]

[i] + [ii] + [iii] + [iv], we get,

AB + BC + BC + CD + CD + DA + DA + AB > CA + DB + AC + DB

2AB + 2BC + 2CD + 2DA > 2CA + 2DB,

2(AB + BC + CA + DA) > 2(CA + DB)

AB + BC + CA + DA > CA + DB

So, the given expression is correct.