Denote the given expression by $S(a, b, c, d)$. We have \[ 1 = \sum_{\textrm{cyc}}\frac{a}{a+b+c+d} < S(a, b, c, d) < \frac{a}{a+b} + \frac{b}{a+b} + \frac{c}{c+d} + \frac{d}{c+d} =2. \] Also, $S(x,x,1,1) \rightarrow 1$ for $x\rightarrow \infty$ and $S(x,1,x,1) \rightarrow 2$ for $x\rightarrow \infty$. Hence the range of $S$ is $]1,2[$.

If $ a,b,c,d$ be positive real variables, let the expression $ S(k) = \frac {a}{d+ka + b} + \frac {b}{a + kb + c} + \frac {c}{b + kc + d} + \frac {d}{ c + kd+a},$ then $ S(k)>\frac{2}{k+1}$ holds if and only if $ 0\leq k\leq 1$; $ S(k)<\frac{2}{k}$ holds if and only if $ 0\leq k\leq 2$.

Was on WOOT. I don't think previous solutions are rigorous enough...(one must show all numbers in the range work). Note that $$\sum_{cyc}\frac{a}{a+b+d}>\sum_{cyc}\frac{a}{a+b+c+d}=1$$Note that $\lim_{(a, b, c, d)\to(\infty, 1, 0, 1)}\sum_{cyc}\frac{a}{a+b+d}=1$, thus the lower bound is $1$, exclusive. Note that $$\sum_{cyc}\frac{a}{a+b+d}<\frac{a}{a+b}+\frac{b}{b+a}+\frac{c}{c+d}+\frac{d}{d+c}=2.$$Note that $\lim_{(a, b, c, d)\to(\infty, 1, \infty, 1)}\sum_{cyc}\frac{a}{a+b+d}=2$, thus the upper bound is $2$, exclusive. Set $(a,b,d)=(\infty,1,1)$, and let $\sum_{cyc}\frac{a}{a+b+d}$ vary with $c$. Clearly our function in terms of $c$ is continuous, thus by the intermediate value theorem, all reals within $(1,2)$ can be attained. Thus $\sum_{cyc}\frac{a}{a+b+d}\in (1,2)$, as desired. $\blacksquare$

Awesome_guy wrote: Was on WOOT. I don't think previous solutions are rigorous enough...(one must show all numbers in the range work). Note that $$\sum_{cyc}\frac{a}{a+b+d}>\sum_{cyc}\frac{a}{a+b+c+d}=1$$Note that $\lim_{(a, b, c, d)\to(\infty, 1, 0, 1)}\sum_{cyc}\frac{a}{a+b+d}=1$, thus the lower bound is $1$, exclusive. Note that $$\sum_{cyc}\frac{a}{a+b+d}<\frac{a}{a+b}+\frac{b}{b+a}+\frac{c}{c+d}+\frac{d}{d+c}=2.$$Note that $\lim_{(a, b, c, d)\to(\infty, 1, \infty, 1)}\sum_{cyc}\frac{a}{a+b+d}=2$, thus the upper bound is $2$, exclusive. Set $(a,b,d)=(\infty,1,1)$, and let $\sum_{cyc}\frac{a}{a+b+d}$ vary with $c$. Clearly our function in terms of $c$ is continuous, thus by the intermediate value theorem, all reals within $(1,2)$ can be attained. Thus $\sum_{cyc}\frac{a}{a+b+d}\in (1,2)$, as desired. $\blacksquare$ I did not get your reasoning when you showed that 1 and 2 are not attained by the function. Could you clarify it, please?