How prove that $|AB - CD| + |AD - BC| \geq 2|AC - BD|$ in cyclic quadrilateral? Let ABCD be a cyclic quadrilateral. How show that $|AB - CD| + |AD - BC| \geq 2|AC - BD|$?
 A: put Circle center on $O$, $ABCD $is counter clockwise , $A(\cos{A},\sin{A}),B(\cos{B},\sin{B})$ etc.
$AB=2|\sin{\dfrac{B-A}{2}}|,CD=2|\sin{\dfrac{D-C}{2}}|$,$|AB-CD|=2||\sin{\dfrac{B-A}{2}}|-|\sin{\dfrac{D-C}{2}}||$
$B-A=2\alpha,C-B=2\beta,D-C=2\gamma \implies \alpha+\beta+\gamma<\pi,0<\alpha,\beta,\gamma<\pi$
the inequality become:
$|\sin{\alpha}-\sin{\gamma}|+|\sin{(\alpha+\beta+\gamma)}-\sin{\beta}|\ge 2|\sin{(\alpha+\beta)}-\sin{(\beta+\gamma)}| \iff |\cos{\dfrac{\alpha+\gamma}{2}}\sin{\dfrac{\alpha-\gamma}{2}}|+|\cos{\dfrac{\alpha+\gamma+2\beta}{2}}\sin{\dfrac{\alpha+\gamma}{2}}| \ge 2 |\cos{\dfrac{\alpha+\gamma+2\beta}{2}}\sin{\dfrac{\alpha-\gamma}{2}}| \iff |\sin{\dfrac{\alpha-\gamma}{2}}|\left(|\cos{\dfrac{\alpha+\gamma}{2}}|-|\cos{\dfrac{\alpha+\gamma+2\beta}{2}}|\right) \ge |\cos{\dfrac{\alpha+\gamma+2\beta}{2}}|\left(|\sin{\dfrac{\alpha-\gamma}{2}}|-|\sin{\dfrac{\alpha+\gamma}{2}}|\right)$
note last one: RHS$\le 0 \iff |\sin{\dfrac{\alpha-\gamma}{2}}|-|\sin{\dfrac{\alpha+\gamma}{2}}| <0,|\cos{\dfrac{\alpha+\gamma+2\beta}{2}}|\ge0$
LHS$\ge 0 \iff |\sin{\dfrac{\alpha-\gamma}{2}}|\ge 0,|\cos{\dfrac{\alpha+\gamma}{2}}|-|\cos{\dfrac{\alpha+\gamma+2\beta}{2}}|=2\sin{\dfrac{\alpha+\gamma+\beta}{2}}\sin{\dfrac{\beta}{2}}>0 $ or $|\cos{\dfrac{\alpha+\gamma}{2}}|-|\cos{\dfrac{\alpha+\gamma+2\beta}{2}}|=2\cos{\dfrac{\alpha+\gamma+\beta}{2}}\cos{\dfrac{\beta}{2}} >0$ 
the "=" only hold when $\alpha=\gamma,\alpha+\gamma+2\beta=\pi$ which means $ABCD$ is a rectangle.
