Question

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

Answer 1

ABCD is a quadrilateral and AC, and BD are the diagonals.
Sum of the two sides of a triangle is greater than the third side.
So, considering the triangle AOB, BOC, COD and AOD, we get
AO + BO > AB
BO + CO > BC
CO + DO > CD
AO + DO > AD

Adding all the above equations, we have

2(AO + BO + CO + DO) > (AB + BC+CD+DA)

⇒ 2[(AO + BO) + (BO + DO)] > (AB + BC+CD+DA)

⇒ 2(AC + BD)>(AB + BC+CD+DA)

Hence, (AB + BC+CD+DA)<2(AC + BD)