Question

in the figure show that 2 ( AC + BD )> AB + BC + CD + D A

Answer 1

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side

Therefore, 

In Δ AOB, AB < OA + OB ……….(i) 

In Δ BOC, BC < OB + OC ……….(ii) 

In Δ COD, CD < OC + OD ……….(iii) 

In Δ AOD, DA < OD + OA ……….(iv) 

⇒ AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD 

⇒ AB + BC + CD + DA < 2[(AO + OC) + (DO + OB)] 

⇒ AB + BC + CD + DA < 2(AC + BD) 

Hence, it is proved.