VII OBM

Let the quadrilateral be $ABCD$. We have $AB+BC>AC$ and $CD+DA>AC$, so the perimeter is bigger than $2\cdot AC$ and, similarly, bigger than $2\cdot BD$. Hence it's bigger than $AC+BD$, and the perimeter minus the sum of the diagonals is bigger than zero. Notice that this inequality is sharp: just consider a rectangle with two sides next to zero.



Now, $\alpha +\beta +\gamma +\delta =180^{\circ }$ and, since the radius of the circle is $1$, $AB=2\sin \gamma$, $BC=2\sin \beta$, $CD=2\sin \alpha$, $AD=2\sin \delta$, $AC=2\sin (\beta +\gamma )=2\sin (\alpha +\delta )=\sin (\beta +\gamma )+\sin (\alpha +\delta )$ and $BD=2\sin (\alpha +\beta )=2\sin (\gamma +\delta )=\sin (\alpha +\beta )+\sin (\gamma +\delta )$, so the difference required is $$2(\sin \alpha +\sin \beta +\sin \gamma +\sin \delta )-\sin (\alpha +\delta )-\sin (\beta +\gamma )-\sin (\alpha +\beta )-\sin (\gamma +\delta )$$ Notice that $\sin (\alpha +\gamma )$ and $\sin (\beta +\delta )$ are “missing”. So we will prove a stronger result, that is, $$\begin{array}{rl}E& =2(\sin \alpha +\sin \beta +\sin \gamma +\sin \delta )\\ & -\sin (\alpha +\delta )-\sin (\beta +\gamma )-\sin (\alpha +\beta )-\sin (\gamma +\delta )-\sin (\alpha +\gamma )-\sin (\beta +\delta )<0\\ ⟹& 2(\sin \alpha +\sin \beta +\sin \gamma +\sin \delta )-\sin (\alpha +\delta )-\sin (\beta +\gamma )-\sin (\alpha +\beta )-\sin (\gamma +\delta )\\ & <\sin (\alpha +\gamma )+\sin (\beta +\delta )\le 2\end{array}$$ First notice that $$\begin{array}{rl}\sin (\alpha +\beta )+\sin (\alpha +\gamma )& =2\sin \left(\alpha +\frac{\beta +\gamma }{2}\right)\cos \left(\frac{\beta -\gamma }{2}\right)\\ & =2\cos \left(\frac{\alpha -\delta }{2}\right)\cos \left(\frac{\beta -\gamma }{2}\right)\end{array}$$ so $$\begin{array}{rl}& \sin \beta +\sin \gamma -\sin (\alpha +\beta )-\sin (\alpha +\gamma )\\ & =2\sin \left(\frac{\beta +\gamma }{2}\right)\cos \left(\frac{\beta -\gamma }{2}\right)-2\cos \left(\frac{\alpha -\delta }{2}\right)\cos \left(\frac{\beta -\gamma }{2}\right)\\ & =2\cos \left(\frac{\beta -\gamma }{2}\right)\left(\cos \left(\frac{\alpha +\delta }{2}\right)-\cos \left(\frac{\alpha -\delta }{2}\right)\right)\\ & =-4\cos \left(\frac{\beta -\gamma }{2}\right)\sin \frac{\alpha }{2}\sin \frac{\delta }{2}\end{array}$$ Summing three analogous equations, we obtain $$\begin{array}{rl}& 2(\sin \alpha +\sin \beta +\sin \gamma )\\ & -\sin (\alpha +\delta )-\sin (\beta +\gamma )-\sin (\alpha +\beta )-\sin (\gamma +\delta )-\sin (\alpha +\gamma )-\sin (\beta +\delta )\\ & =-4\sin \frac{\delta }{2}\left(\cos \left(\frac{\beta -\gamma }{2}\right)\sin \frac{\alpha }{2}+\cos \left(\frac{\alpha -\gamma }{2}\right)\sin \frac{\beta }{2}+\cos \left(\frac{\beta -\alpha }{2}\right)\sin \frac{\gamma }{2}\right)\\ & =-4\sin \frac{\delta }{2}\left(\sin \left(\frac{\alpha +\beta -\gamma }{2}\right)+\sin \left(\frac{\alpha -\beta +\gamma }{2}\right)+\sin \left(\frac{-\alpha +\beta +\gamma }{2}\right)\right)\end{array}$$ Finally, $$\begin{array}{rl}E& =2\sin \delta -4\sin \frac{\delta }{2}\left(\sin \left(\frac{\alpha +\beta -\gamma }{2}\right)+\sin \left(\frac{\alpha -\beta +\gamma }{2}\right)+\sin \left(\frac{-\alpha +\beta +\gamma }{2}\right)\right)\\ & =-4\sin \frac{\delta }{2}\left(-\cos \frac{\delta }{2}+\sin \left(\frac{\alpha +\beta -\gamma }{2}\right)+\sin \left(\frac{\alpha -\beta +\gamma }{2}\right)+\sin \left(\frac{-\alpha +\beta +\gamma }{2}\right)\right)\\ & =-4\sin \frac{\delta }{2}\left(-\sin \left(\frac{\alpha +\beta +\gamma }{2}\right)+\sin \left(\frac{\alpha +\beta -\gamma }{2}\right)+2\cos \left(\frac{\alpha -\beta }{2}\right)\sin \frac{\gamma }{2}\right)\\ & =-8\sin \frac{\delta }{2}\left(-\cos \left(\frac{\alpha +\beta }{2}\right)\sin \frac{\gamma }{2}+\cos \left(\frac{\alpha -\beta }{2}\right)\sin \frac{\gamma }{2}\right)\\ & =-8\sin \frac{\gamma }{2}\sin \frac{\delta }{2}\left(-\cos \left(\frac{\alpha +\beta }{2}\right)+\cos \left(\frac{\alpha -\beta }{2}\right)\right)\\ & =-16\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\sin \frac{\gamma }{2}\sin \frac{\delta }{2}<0\end{array}$$ because $0<\frac{\alpha }{2},\frac{\beta }{2},\frac{\gamma }{2},\frac{\delta }{2}<90^{\circ }$.