Question

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

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 $\Delta$AOB, AB < OA+OB ……….(i)
In $\Delta$BOC, BC < OB+OC ……….(ii)
In $\Delta$COD, CD < OC+OD ……….(iii)
In $\Delta$AOD, DA < OD+OA ……….(iv)
Adding eq. (i), (ii), (iii) and(iv),
$\Rightarrow$ AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD +OA
$\Rightarrow$ AB + BC + CD + DA < 2OA + 2OB + 2OC +2OD
$\Rightarrow$ AB + BC + CD + DA < 2[(AO + OC) + (DO +OB)]
$\Rightarrow$ AB + BC + CD + DA < 2(AC +BD)
Hence, it is proved.