5. ABCD is quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)
We consider ABCD as quadrilateral and P is the point where the diagonals intersect. As shown in the figure below.
We know that the sum of the length of any two sides is always greater than the third side.
Now we consider the ΔPAB,
Here, PA + PB > AB … [equation i]
Then, we consider the ΔPBC
Here, PB + PC > BC … [equation ii]
Now, we consider the ΔPCD
Here, PC + PD > CD … [equation iii]
Now, we consider the ΔPDA
Here, PD + PA > DA … [equation iv]
Adding equation [i], [ii], [iii] and [iv] we get,
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) > AB + BC + CD + DA
From the figure we have, AC = PA + PC and BD = PB + PD
2AC + 2BD > AB + BC + CD + DA
2(AC + BD) > AB + BC + CD + DA
Hence, the given expression is true.