Question

In a quadrilateral $ABCD$; prove that:
$AB+BC+CD+DA>2BD$.

Answer 1

Consider Quadrilateral $ABCD$

In $△ABD,$ From Triangle Inequality,

$AB+AD>BD\cdots \left(1\right)$

In $△BCD,$ From Triangle Inequality,

$BC+DC>BD\cdots \left(2\right)$

Adding both the equations

$⟹AB+BC+CD+AD>2BD$