Question

In a quadrilateral $ABCD$; prove that:
$AB+BC+CD+AD>AC+BD$

Answer 1

Now by property of triangles we know that sum of any two sides $>$ third side. Therefore $AB+BC>AC$ same way $CD+DA>AC$ similarly way we can get $DA+AB>BD$ and $BC+CD>BD$

Now, adding all inequalities $2(AB+BC+CD+DA)>2(AC+BD)$

Dividing both sides by $2$ we get

$\Rightarrow (AB+BC+CD+DA)>(AC+BD)$

Hence, proved.