Let

S(a ,b ,c ) = a/(b+c) + b/(a+c) + c/(a+b)

and

C = { n ≥ 1 : ∃ a ,b ,c ≥ 1  S(a ,b ,c ) = n } .

1. The problem is equivalent to a single polynomial equation

   F(a ,b ,c ,n) = 0 ,

   where F is obtained by clearing the denominators:

   a(a+b)(a+c) + b(a+b)(b+c) + c(a+c)(b+c)
   

   – n (a+b)(a+c)(b+c) = 0 .

   Expanding, F is an integral polynomial of total degree 3 in the variables a ,b ,c (and of degree 1 in n, so of total degree 4 if n is counted as a variable).

2. Computably-enumerable (r.e.) but not obviously decidable

   Because the variables range over the positive integers, we can systematically enumerate all triples (a ,b ,c) and record every value n for which F(a ,b ,c ,n)=0. Hence C is r.e.: there is a Turing machine that lists its members.

   What we do not know is whether C is co-r.e. as well, i.e. whether there is an algorithm that is guaranteed to say “no” after finitely many steps when n∉C. That is exactly the question you asked: is C computable (decidable)?

3. Relation to Hilbert’s Tenth Problem with bounded degree

   Matiyasevich’s negative solution of Hilbert’s Tenth Problem shows that the general existential theory of the integers (∃ x₁,…,x_k : P=0) is undecidable. Moreover, Jones, Sato, Wada and Wiens later produced a universal polynomial of total degree 4 that represents every r.e. set. Hence membership in a set like C can certainly be undecidable when degree 4 is allowed.

   The crucial point here is that our polynomial F has total degree 3 in the quantified (a ,b ,c) variables. Whether degree 3 already suffices to obtain undecidability is a famous open problem. No one presently knows whether every r.e. set can be represented by a cubic Diophantine equation, nor whether the solvability of a single fixed cubic equation can be undecidable.

   Therefore:

   • If one could prove that C is not computable, one would at the same time prove that a single cubic equation can encode an undecidable set—settling that long-standing open problem in the negative direction.

   • If one could prove that C is computable, this would give a non-trivial algorithmic decidability result for a class of cubic equations that presently lies far beyond what is known (compare with the decidability of Thue equations in two variables, but degree ≥ 3 and four variables is wide open).

4. State of the art

   No algorithm is known that decides C, and no proof of undecidability is known either. In the literature on Diophantine definability this particular equation has not been isolated before, but it fits squarely into the still unresolved “degree-3 frontier”.

5. Summary

   • C is recursively enumerable.
   • Whether C is decidable is currently unknown.
   • A resolution in either direction would solve a well-known open problem about the power of cubic Diophantine equations.

So the best answer at present is: we do not know; the question is open.