Bounds on $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}$

Question

It's simple. What are the bounds on $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}$ as $a,b,c>0$. Thanks in advance!

Edit: I need bound that can actually be touched i.e. $\alpha$ and $\beta$ such that $\alpha\leq\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\leq\beta$

Answer 1

If $a,\,b,\,c$ are positive real numbers then the answer is $2.$

Indeed, we have $$\frac{c+a}{a+b+c}-\frac{a}{a+b}=\frac{bc}{(a+b)(a+b+c)} \geqslant 0,$$ so $$\frac{a}{a+b} \leqslant \frac{c+a}{a+b+c},$$ therefore $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \leqslant \frac{(c+a)+(a+b)+(b+c)}{a+b+c}=2.$$ In addition, from $$\frac{a}{a+b}>\frac{a}{a+b+c},$$ we get $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>1.$$

Answer 2

There are no well defined bounds.

Let $S$ denote the given expression. Note first that $$S > \frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1.$$ On the other hand, with $a=1$, $c=\varepsilon$ and $b=\varepsilon^2$ we have $$S=\frac{1}{1+\varepsilon^2} + \frac{2\varepsilon}{1+\varepsilon} \to 1$$ as $\varepsilon \downarrow 0$. Thus the largest lower bound for $S$ is $1$. For an upper bound, assume wlog that $a \leq b,c$. Then $$S \leq \frac{a}{a+b} + \frac{b}{b+a}+\frac{c}{c+a} < 2.$$ With $a=1$, $b=\varepsilon$ and $c=\varepsilon^2$ we have $$S = \frac{2}{1+\varepsilon} + \frac{\varepsilon^2}{1+\varepsilon^2} \to 2$$ as $\varepsilon \downarrow 0$. Hence the smallest upper bound for $S$ is $2$.