Knowing that $a,b,c \in ℝ^*_+$ prove that $\frac{a+b}{a+b-c},\frac{b+c}{b+c-a},\frac{c+a}{c+a-b} $ don't belong simultaneously to the interval $(1,2)$

Question

I have to solve the following problem but I don't know how to :

Knowing that $a,b,c \in ℝ^*_+$ prove that $\frac{a+b}{a+b-c},\frac{b+c}{b+c-a},\frac{c+a}{c+a-b} $ don't belong simultaneously to the interval $(1,2).$

Here's what I've tried:

I worked on cases $$\begin{cases} a \ge b \ge c \vee a \ge c \ge b \rightarrow \frac{a+b}{b+c-a}<0 \\ b \ge a \ge c \vee b \ge c \ge a \rightarrow \frac{c+a}{c+a-b}<0 \\ c \ge a \ge b \vee c \ge b \ge a \rightarrow \frac{a+b}{a+b-c}<0 \end{cases} $$ Therefore $a,b,c$ can't be simultaneously in the interval $(1,2)$

Answer 1

Assume that they all belong to $(1,2)$ then $$ \frac{a+b-c}{a+b},\quad\frac{a-b+c}{a+c},\quad \frac{-a+b+c}{b+c} $$ all belong to $\left(\frac{1}{2},1\right)$, hence $$ \frac{c}{a+b},\quad \frac{b}{a+c},\quad \frac{a}{b+c} $$ all belong to $\left(0,\frac{1}{2}\right)$ and $$ 2c<(a+b),\quad 2b<(a+c),\quad 2a<(b+c). $$ The last three inequalities cannot hold at the same time, since $$ 2c+2b+2a = (a+b)+(a+c)+(b+c).$$

Answer 2

Assume $a\leq b\leq c$

Notice $\frac{a+b}{a+b-c}\in(1,2)\implies \frac{a+b-c}{a+b}\in(\frac{1}{2},1)\implies \frac{c}{a+b}\in(0,\frac{1}{2})$, but $\frac{c}{a+b}\geq\frac{c}{2c}=\frac{1}{2}$