MathDB

a,b,>1: x < 1, y < 1, ax+by < 1, 1/(1-ax-by) \le a/(1-x)+b/(1-y) <=> a+b = 1

Source: Germany 1996 p2

February 22, 2020

inequalitiesalgebra

Problem Statement

Let

a

and

b

be positive real numbers smaller than

1

. Prove that the following two statements are equivalent: (i)

a+b = 1

, (ii) Whenever

x,y

are positive real numbers such that

x < 1, y < 1, ax+by < 1

, the following inequlity holds:

\frac{1}{1-ax-by} \le \frac{a}{1-x} + \frac{b}{1-y}