Question

In quadrilateral ABCD prove AB+BC+CD+DA is greater than 2AC

Answer 1

Sum of two sides in a triangle is greater than the third side
Apply it in two triangles formed by diagonal AC and add those two inequalities
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Answer 2

Answer:

In ∆ ABC

Ab + bc > ac -- 1 (sum of two sides is greater than the third side.)

In ∆ adc

Cd + Da > ac --2 (sum of two sides is greater than the third side)

On adding 1 and 2 equation, we get

Cd + Da + ab + bc > 2ac

Hope it helps