Prove AB CD + AD BC < AC(AB+AD) for Convex Quadrilateral ABCD

[anemone]

Here is this week's POTW:

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The sum of the angles $A$ and $C$ of a convex quadrilateral $ABCD$ is less than $180^\circ$.

Prove that $AB\cdot CD+AD \cdot BC <AC(AB+AD)$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

No one answered last week's problem.(Sadface)

You can find the suggested solution below:

Spoiler

Let $O$ be the circumcircle $ABD$. Then the point $C$ is outside this circle, but inside the $\angle BAD$. Let us apply the inversion with the center $A$ and radius $1$. This inversion maps the circle $O$ to the line $O'=B'D'$, where $B'$ and $D'$ are images of $B$ and $D$. The point $C$ goes to the point $C'$ inside the triangle $AB'D'$. Therefore $B'C'+C'D'<AB'+AD'$. Now, due to inversion properties, we have

$B'C'=\dfrac{BC}{AB\cdot AC},\,C'D'=\dfrac{CD}{AC\cdot AD},\,AB'=\dfrac{1}{AB},\,AD'=\dfrac{1}{AD}$

Hence

$\dfrac{BC}{AB\cdot AC}+\dfrac{CD}{AC\cdot AD}<\dfrac{1}{AB}+\dfrac{1}{AD}$

Multiplying the above by $AB\cdot AC \cdot AD$ the result follows.