IMO 1991 SL 6

Origin: USS

Problem

Prove for each triangle ABC the inequality 4 < IA \cdot IB \cdot IC lAlBlC \leq8 27, where I is the incenter and lA, lB, lC are the lengths of the angle bisectors of ABC.

Solution

Let a, b, c be sides of the triangle. Let A1 be the intersection of line AI with BC. By the known fact, BA1 : A1C = c : b and AI : IA1 = AB : BA1, hence BA1 = ac b+c and AI IA1 = AB BA1 = b+c a . Consequently AI lA = b+c a+b+c. Put a = n+p, b = p+m, c = m+n: it is obvious that m, n, p are positive. Our inequality becomes 2 < (2m + n + p)(m + 2n + p)(m + n + 2p) (m + n + p)3 \leq64 27. The right side inequality immediately follows from the inequality between arithmetic and geometric means applied on 2m + n + p, m + 2n + p and m+n+2p. For the left side inequality, denote by T = m+n+p. Then we can write (2m + n + p)(m + 2n + p)(m + n + 2p) = (T + m)(T + n)(T + p) and (T +m)(T +n)(T +p) = T 3+(m+n+p)T 2+(mn+np+pn)T +mnp > 2T 3. Remark. The inequalities cannot be improved. In fact, AI\cdotBI\cdotCI lAlBlC is equal to 8/27 for a = b = c, while it can be arbitrarily close to 1/4 if a = b and c is sufficiently small.