MathDB
\frac{a}{a+c} + \frac{b}{b+a} > \frac{c}{b+c} , ax^2 - 4bx + 4c = 0

Source: - 2001 Cuba MO 2.6

September 15, 2024
algebrainequalities

Problem Statement

The roots of the equation ax2−4bx+4c=0ax^2 - 4bx + 4c = 0 with a>0 a > 0 belong to interval [2,3][2, 3]. Prove that: a) a≤b≤c<a+b.a \le b \le c < a + b. b) aa+c+bb+a>cb+c.\frac{a}{a+c} + \frac{b}{b+a} > \frac{c}{b+c} .
Back to ProblemsView on AoPS