Mathematical Reflections 2016, Issue 3 - Problem S374

Problem:
Let $a,b,c$ be positive real numbers. Prove that at least one of the numbers $$\dfrac{a+b}{a+b-c}, \qquad \dfrac{b+c}{b+c-a}, \qquad \dfrac{c+a}{c+a-b}$$ is not in the interval $(1,2)$.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Assume by contradiction that all the given numbers are in the interval $(1,2)$. Then,
$$\dfrac{a+b}{a+b-c} > 1, \qquad \dfrac{b+c}{b+c-a} > 1, \qquad \dfrac{c+a}{c+a-b} > 1,$$ which gives
$$\dfrac{c}{a+b-c}>0, \qquad \dfrac{a}{b+c-a}>0, \qquad \dfrac{b}{c+a-b}>0.$$ Since $a,b,c$ are positive real numbers, then $a+b-c>0, b+c-a>0$ and $c+a-b>0$. Moreover,
$$\dfrac{a+b}{a+b-c} < 2, \qquad \dfrac{b+c}{b+c-a} < 2, \qquad \dfrac{c+a}{c+a-b} < 2,$$ which gives $$\dfrac{2c-a-b}{a+b-c}<0, \qquad \dfrac{2a-b-c}{b+c-a}<0, \qquad \dfrac{2b-c-a}{c+a-b}<0.$$ Since all the denominators are positive, then $2c-a-b<0$, $2a-b-c<0$, $2b-c-a<0$. Adding these three inequalities, we get $0<0$, contradiction. It follows that one of the three given numbers is not in the interval $(1,2)$.