50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010)

Question

In a convex quadrilateral ABCD angles ABC and BCD are not less than 120^. Prove, that AC + BD > AB + BC + CD. FIGURE_0

Accepted Answer

Let AB = a, BC = b, CD = c (Fig.18). Then, AC^2 = a^2 + b^2 - 2ab B a^2 + ab + b^2. By analogy, BD^2 b^2 + bc + c^2 AC + BD a^2 + ab + b^2 + b^2 + bc + c^2. Since a^2 + ab + b^2 > a + 12b and b^2 + bc + c^2 > c + 12b, then AC + BD > a + b + c = AB + BC + CD. FIGURE_0 Fig.18