Each sidelength of a convex quadrilateral $ABCD$ is not less than $1$ and not greater than $2$. The diagonals of this quadrilateral meet at point $O$. Prove that $S_{AOB}+ S_{COD} \le 2(S_{AOD}+ S_{BOC})$.
Source: Sharygin 2013 Final 9.3
Tags: geometry, Inequality, areas
Each sidelength of a convex quadrilateral $ABCD$ is not less than $1$ and not greater than $2$. The diagonals of this quadrilateral meet at point $O$. Prove that $S_{AOB}+ S_{COD} \le 2(S_{AOD}+ S_{BOC})$.