Question

In a quadrilateral $ABCD$, show that $(AB+BC+CD+DA)<2(AC+BD)$

Answer 1

Consider $△AOB$

We know that

$AO+BO>AB..\left(1\right)$

Consider $△BOC$

We know that

$BO+CO>BC..\left(2\right)$

Consider $△COD$

We know that

$CO+DO>CD..\left(3\right)$

Consider $△AOD$

We know that

$DO+AO>DA..\left(4\right)$

By adding all the equations

$AO+BO+BO+CO+CO+DO+DO+AO>AB+BC+CD+DA$

So we get

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

On further calculation

$2AC+2BD>AB+BC+CD+DA$

By taking $2$ as common

$2(AC+BD)>AB+BC+CD+DA$

So we get

$AB+BC+CD+DA<2(AC+BD)$

Therefore, it is proved that $AB+BC+CD+DA<2(AC+BD)$.