Can you prove the result? If stuck, click the Hint in the sketch.

Can you provide counter-examples for the result when ABCD is concave or crossed?

(The problem was used in the Senior 3rd Round of the South African Mathematics Olympiad (SAMO) in 2001 as the first 'warm-up' problem. A solution is given at SAMO Solutions: 2001. Past SAMO papers & solutions for several years are available at Past SAMO Papers & Solutions.)

Some Applications^{1}

(1) Using the above inequality together with a familiar property of the sum of the opposite sides of a circumscribed quadrilateral ABCD, it follows immediately that for a convex, circumscribed ABCD the sum of the diagonals divided by the sum of any pair of opposite sides is always greater than 1 and less than 2. For example: 1 < (AC + BD)/(AB + CD) < 2 as shown below.

Circumscribed Quadrilateral Inequality

What do you notice about the maximum value of the ratio (AC + BD)/(AB + CD)? When does it occur and what is the precise value of its upper limit? Can you explain (prove) your observation?

(2) Actually, the lower bound of the above inequality for a convex, circumscribed quadrilateral applies to any convex quadrilateral! In other words, the sum of any pair of opposite sides of a convex quadrilateral is always less than the sum of the diagonals as shown below.

Another convex quadrilateral inequality

Can you explain (prove) why this result is true?

Footnote^{1} These two simple, but interesting applications were discovered by the author on 8 Dec 2010 upon reflecting on possible implications/applications of the quadrilateral diagonal-perimeter inequality at the top.

Another Quadrilateral Inequality

One of the most famous quadrilateral inequalities is that of Ptolemy (~ 100-168 AD), which states that for any quadrilateral ABCD, AB*CD + BC*AD >= AC*BD (and equality only holds when ABCD is cyclic (& convex)). Read more about the result and its applications at Ptolemy's Theorem.