A hint on finding all solutions of $S = \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$? I've been working at this for awhile but I haven't been able to figure out the right approach. The question is to find all values of $S = \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$ for $a,b,c,d > 0$. Anyone have a hint (no solutions, please) at a more principled way to approach the problem?
I've been just throwing stuff at the wall and trying to find something that sticks.
What I've tried so far:
Plugging in some quick values. Putting $a=b=c=d=1$ yields $S = 4/3$. Taking $a=c=1, b=d=2$ yields $7/5$ so $4/3$ is not the only answer. Note: if $(a,b,c,d)$ yields $k$ then $(sa,sb,sc,sd)$ yields $k$ for any $s > 0$.
If I fix $a=b=c=1$ then $S = 4/3$ regardless of the value of $d$.
If I fix $a=c, b=d$, I can write $b = ka$, which gives $\frac{2}{2k+1} + \frac{2k}{k+2} = S$. This is a quadratic in $k$, with positive discriminant for any value of $S$. But $k$ is not necessarily positive, so I can't claim that all values of $S$ are valid.
Getting some quick bounds: $S > \frac{a}{a+b+c+d} + \frac{b}{a+b+c+d} + \frac{c}{a+b+c+d} + \frac{d}{a+b+c+d} = 1$, and $S < a/a + b/b + c/c + d/d = 4$, so I know $1 < S < 4$. The thought is to either try and tighten these bounds somehow (not sure how to approach this) or figure out how to represent an arbitrary $x$ in this range with some choice of $a,b,c,d$.
Some substitutions: Imposing the constraint $a+b+c+d = 1$ or $u = a+b+d, v = a+b+c, w = b+c+d, x = a+c+d$ come to mind, both attempts at removing the sums from the denominator. Respectively those give $S = \frac{a}{1-c} + \frac{b}{1-d} + \frac{c}{1-a} + \frac{d}{1-b}$ and $S = \frac{u+v+w-2x}{3u} + \frac{u+v+w-2x}{3v} + \frac{v+x+w-2u}{3w} + \frac{u+w+x-2v}{3x}$.
Another way of looking at things: $\frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$ is continuous in $(a,b,c,d)$. And I think we can get arbitrarily close to 1 by setting $a$ very large, and then $c$ very small relative to $b + d$. So 1 is probably the $\inf$ of $S$. If we do the opposite and set $a$ large and $c$ large, we can get close to 2. My suspicion is that $2$ is the $\sup$ but I'm not sure how to prove it.
 A: For $c=d\rightarrow0^+$ we see that $$S\rightarrow\frac{a}{a+b}+\frac{b}{a+b}=1$$ and
$$\sum_{cyc}\frac{a}{a+b+d}>\sum_{cyc}\frac{a}{a+b+c+d}=1.$$
For $b=d\rightarrow0^+$ we see that $S$  close to $2$.
We'll prove that $$\sum_{cyc}\frac{a}{a+b+d}<2.$$
Indeed, we need to prove that
$$\sum_{cyc}\left(\frac{a}{a+b+d}-1\right)<-2$$ or
$$\sum_{cyc}\frac{b+d}{a+b+d}>2,$$ which is true by C-S:
$$\sum_{cyc}\frac{b+d}{a+b+d}=\sum_{cyc}\frac{(b+d)^2}{(b+d)(a+b+d)}\geq\frac{4(a+b+c+d)^2}{\sum\limits_{cyc}(b+d)(a+b+d)}=$$
$$=\frac{4(a+b+c+d)^2}{\sum\limits_{cyc}(2ab+2a^2+2ac)}=\frac{2(a+b+c+d)^2}{\sum\limits_{cyc}(a^2+ab+ac)}>2.$$
Since $S$ is a continuous function, we got the best estimations:
$$1<S<2.$$
Another way to get $$\sum_{cyc}\frac{b+d}{a+b+d}>2.$$
$$\sum_{cyc}\frac{b+d}{a+b+d}=(b+d)\left(\tfrac{1}{a+b+d}+\tfrac{1}{c+d+b}\right)+(a+c)\left(\tfrac{1}{a+b+c}+\tfrac{1}{a+c+d}\right)>$$
$$>(b+d)\left(\tfrac{1}{a+b+c+d}+\tfrac{1}{a+b+c+d}\right)+(a+c)\left(\tfrac{1}{a+b+c+d}+\tfrac{1}{a+b+c+d}\right)=2$$
A: Because
$$\frac{a}{a+b+d}>\frac{a}{a+b+c+d},$$
so
$$S>\frac{a+b+c+d}{a+b+c+d}=1.$$
And
$$\frac{c+a}{a+b+c+d}-\frac{a}{a+b+d} = \frac{c(b+d)}{(a+b+c+d)(a+b+d)}>0,$$
so
$$\frac{a}{a+b+d}<\frac{c+a}{a+b+c+d},$$
therefore
$$S<\frac{2(c+a)+2(b+d)}{a+b+c+d}=2.$$
We get the estimations $1<S<2.$
A: We have
$$\frac{a}{a+b+d} < \frac{a}{a+b}, \frac{b}{a+b+c} < \frac{b}{a+b}, \frac{c}{b+c+d} < \frac{c}{c+d}, \frac{d}{a+c+d} < \frac{d}{c+d}$$
So $S < \frac{a+b}{a+b} + \frac{c+d}{c+d} = 2$, which gives the upper bound that I was missing. Then as mentioned in the question, send $a$, $c$ to be large, (or $b,d\to 0$ as in Michael's answer) to get arbitrarily close to 2, which gives the missing piece.
The case of $S < 1$ is also covered in the question.
