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

Make ABCD a quadrilateral, and mark a point P where the diagonals intersect. As shown in the figure below.

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 ΔPAB,

 AB < PA + PB  … [i]

Now in ΔPBC

BC < PB + PC  … [ii]

Now in ΔPCD

CD < PC + PD  … [iii]

And in ΔPDA

 DA < PD + PA … [iv]

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

AB + BC + CD + DA < PA + PB + PB + PC + PC + PD + PD + PA

AB + BC + CD + DA < 2PA + 2PB + 2PC + 2PD 

 AB + BC + CD + DA < 2PA + 2PC + 2PB + 2PD 

 AB + BC + CD + DA < 2(PA + PC) + 2(PB + PD) 

As we can see from the figure, AC = PA + PC and BD = PB + PD

So,

AB + BC + CD + DA < 2AC + 2BD 

AB + BC + CD + DA < 2(AC + BD) 

So, the given expression is correct.