MathDB
Greater or equal than n / (x+y)

Source: Vietnam TST 1991 for the 32nd IMO, problem 2

June 25, 2005
inequalitiesinequalities unsolved

Problem Statement

For a positive integer n>2 n>2, let (a1,a2,,an) \left(a_{1}, a_{2}, \ldots, a_{n}\right) be a sequence of n n positive reals which is either non-decreasing (this means, we have a1a2an a_{1}\leq a_{2}\leq \ldots \leq a_{n}) or non-increasing (this means, we have a1a2an a_{1}\geq a_{2}\geq \ldots \geq a_{n}), and which satisfies a1an a_{1}\neq a_{n}. Let x x and y y be positive reals satisfying xya1a2a1an \frac{x}{y}\geq \frac{a_{1}-a_{2}}{a_{1}-a_{n}}. Show that: a1a2x+a3y+a2a3x+a4y++an1anx+a1y+ana1x+a2ynx+y. \frac{a_{1}}{a_{2}\cdot x+a_{3}\cdot y}+\frac{a_{2}}{a_{3}\cdot x+a_{4}\cdot y}+\ldots+\frac{a_{n-1}}{a_{n}\cdot x+a_{1}\cdot y}+\frac{a_{n}}{a_{1}\cdot x+a_{2}\cdot y}\geq \frac{n}{x+y}.