Proving Inequalities for Cyclic Quadrilaterals

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SUMMARY

The discussion centers on proving the inequality |AB - CD| + |AD - BC| ≥ 2|AC - BD| for a convex cyclic quadrilateral ABCD. Participants confirm that the convexity condition is redundant since cyclic quadrilaterals are inherently convex. The conversation highlights the use of triangle inequalities derived from the triangles ABC, BCD, ACD, and ABD, although the attempt to manipulate these inequalities to match the problem statement did not yield success.

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Homework Statement


Let ABCD be a convex cyclic quadrilateral. Prove that

|AB-CD|+|AD-BC| \geq 2|AC-BD|

Homework Equations


The Attempt at a Solution


First, isn't a cyclic quadrilateral always convex?

http://en.wikipedia.org/wiki/Cyclic_quadrilateral
 
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ehrenfest said:
First, isn't a cyclic quadrilateral always convex?

Hi ehrenfest! :smile:

Yes … "convex" seems unnecessary!
 
Putnam is supposed to be for fun Ehrenfest. If you ask for help on every problem that you can't immediately solve... how are you having fun? The pleasure is all in finding the aha! moment yourself.
 
tiny-tim said:
Hi ehrenfest! :smile:

Yes … "convex" seems unnecessary!

So, I can get triangle inequalities for the triangles ABC, BCD, ACD, ABD. Put there are 12 of them and I tried to play around with so they would look similar to the inequality in the problem statement but I did not get very far.
 

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