IMO 1997 SL 7

Origin: RUS

Problem

Let ABCDEF be a convex hexagon such that AB = BC, CD = DE, EF = FA. Prove that BC BE + DE DA + FA FC \geq3 2. When does equality occur?

Solution

Let us set AC = a, CE = b, EA = c. Applying Ptolemy’s inequality for the quadrilateral ACEF we get AC \cdot EF + CE \cdot AF \geqAE \cdot CF. Since EF = AF, this implies F A F C \geq c a+b. Similarly BC BE \geq a b+c and DE DA \geq b c+a. Now, BC BE + DE DA + FA FC \geq a b + c + b c + a + c a + b. Hence it is enough to prove that a b + c + b c + a + c a + b \geq3 2. (1) If we now substitute x = b + c, y = c + a, z = a + b and S = a + b + c the inequality (1) becomes equivalent to S(1/x + 1/y + 1/y) −3 \geq3/2 which follows immediately form 1/x + 1/y + 1/z \geq9/(x + y + z) = 9/(2S). Equality occurs if it holds in Ptolemy’s inequalities and also a = b = c. The former happens if and only if the hexagon is cyclic. Hence the only case of equality is when ABCDEF is regular.