Question

In a quadrilateral $ABCD$ prove that:
$AB+BC+CD+DA>2AC$

Answer 1

Consider Quadrilateral $ABCD$

In $△ABC,$ From Triangle Inequality,

$AB+BC>AC\cdots \left(1\right)$

In $△ADC,$ From Triangle Inequality,

$AD+DC>AC\cdots \left(2\right)$

Adding both th equations

$⟹AB+BC+CD+AD>2AC$