MathDB
f(x^2-1998x)-f^2(2x-1999)>=1/4 (VI Soros Olympiad 1990-00 R1 9.4)

Source:

May 27, 2024
algebrafunctionalfunctional inequationinequalities

Problem Statement

Is there a function f(x)f(x), which satisfies both of the following conditions:
a) if x≠yx \ne y, then f(x)≠f(y)f(x)\ne f(y)
b) for all real xx, holds the inequality f(x2−1998x)−f2(2x−1999)≥14f(x^2-1998x)-f^2(2x-1999)\ge \frac14?
Back to ProblemsView on AoPS