Minimum and maximum with $1+\frac{a+x}{b+x}+\frac{a+x}{c+x}$ Let $a,b,c\geq 1$ and $x,y,z\geq 0$. What are the minimum and maximum of
$$f(a,b,c,x,y,z)=\frac{1}{1+\frac{a+x}{b+x}+\frac{a+x}{c+x}}+\frac{1}{1+\frac{b+y}{a+y}+\frac{b+y}{c+y}}+\frac{1}{1+\frac{c+z}{a+z}+\frac{c+z}{b+z}}?$$
Since each term is no more than $1$, the maximum is not more than $3$. When $a=b=c$, $f(a,b,c,x,y,z)=1$. 

Without loss of generality, take $a\ge b\ge c$.
By mediant inquality,
$$1 \le \frac{a+x}{b+x} \le \frac{a}{b}$$
$$1 \le \frac{a+x}{c+x} \le \frac{a}{c}$$

$$3 \le 1+\frac{a+x}{c+x}+\frac{a+x}{c+x} \le
 1+\frac{a}{b}+\frac{a}{c}$$

$$\frac{b}{a} \le \frac{b+y}{a+y} \le 1$$
$$1 \le \frac{b+y}{c+y} \le \frac{b}{c}$$

$$2+\frac{b}{a} \le 1+\frac{b+y}{a+y}+\frac{b+y}{c+y} \le 2+\frac{b}{c}$$

By mediant inquality,
$$\frac{c}{a} \le \frac{c+z}{a+z} \le 1$$
$$\frac{c}{b} \le \frac{c+z}{b+z} \le 1$$

$$1+\frac{c}{a}+\frac{c}{b} \le 1+\frac{c+z}{a+z}+\frac{c+z}{b+z} \le 3$$

For fixed $a\ge b\ge c$,
$$\frac{1}{1+\frac{a}{b}+\frac{a}{c}}+\frac{1}{2+\frac{b}{c}}+\frac{1}{3}
 \le f(a,b,c,x,y,z) \le \frac{1}{3}+\frac{1}{2+\frac{b}{a}}+\frac{1}{1+\frac{c}{a}+\frac{c}{b}}
$$
Take $\frac{a}{b}, \frac{a}{c}, \frac{b}{c} \to \infty$,
$$\frac{1}{3} \le f(a,b,c,x,y,z) \le \frac{11}{6}$$
