MathDB
(a+c)/(b+c) > a/b - [Iran Second Round 1985]

Source:

December 29, 2010
inequalitiesinequalities proposed

Problem Statement

Let a,ba,b and cc be real numbers with b,c>0.b,c >0. Prove that if a<b (a>b), a<b \ ( a>b), then a+cb+c>ab(a+cb+c<ab)\frac{a+c}{b+c} > \frac ab \qquad ( \frac{a+c}{b+c} < \frac ab) And then prove that a+cb+c\frac{a+c}{b+c} is between 11 and ab.\frac ab.