I rear that the formulation of the question is not really good.

"separate $G$ into two edge-disjoint spanning subgraphs"

separate how? separate the vertices into two sets? because then the spanning subgraphs will be obviously edge-disjoint. (so why is that clausa?) also, a cut will obviously separate the graph into two graphs both connected..

separate by edges? but then there are no spanning subgraphs?

please, state your question clearly!

Dear Bogdán Zaválnij,

Thank you for your attention!

Yes,  "sepatate" maybe the wrong word by the definition of Graph theory. Sorry for that.

I mean "separate"  the  edges into to parts, and "separate" the degree of vertices also.

That is to say, if a subgraph   contain the edge $e$, then I consider it contain the vertices of $e$.

Dear Patrice Ossona de Mendez，

 Thank you for your answer!

I need a lower bound than $n$.

It is not hard to prove that every simple graph G (with no multiple edges), which is not necessarily connected, with minimum degree >= 3, can be decomposed to G1 and G2 such that f(G1)+f(G2) =4, then removing edge between them yeilds a smaller counterexample, which contradicts to the minimality of G. So we can assume that any two vertices of degree >=4 are non-adjacent.

If D

Dear Aleksey Nikolaevich Glebov,

Sorry, I think your answer is different with what I want . Nash-Williams's forest cover is different with my "separate" in sense that forest may contain trivial tree.

Do you mean by trivial a tree with just one vertex? If so, then I think it is easy to settle by slightly modifying forests F1 and F2. I will give a more detailed answer a bit later (today or tomorrow).

well, I think there is still some unclearness on the word" separate". (as far I would like to understand it, I still cannot, and the "spanning" is not compatible with it.

so please, please, give us a formal definition of the problem.  because each of us would understand it differently and provide an answer which is not satisfactory to you. (I also can give such an answer, because I see a solution where f(G1)+f(G2)=2 which is clearly

My guess about the statement of the problem is as follows. Given a simple graph G = (V,E) with minimum degree >= 3 find a partition of its edges E = E1 U E2 such that the subgraphs G1 = (V,E1) and G2 = (V,E2) have no isolated vertices (i.e. have minimum degree >=1) and the value f(G1)+f(G2) is as small as possible. Here G1 and G2 are "spanning" in the sence that each of them has the same vertex set as G and they ""separate" the degree of vertices of G" because each vertex of G is incident with at least one edge from both E1 and E2. My previous solution does not satisfy the second requirement because forests F1 and F2 in Nash-Williams Theorem can have isolated vertices. So, my next goal is to show that we can modify F1 and F2 in such a way that each of them have minimum degree 1.  

Well, by my previous solution, if maximum degree D(G)=D 0 the edge Vi-1Vi is coloured 2 while all other edges at Vi are coloured 1 (otherwise, we can modify F1 and F2 by switching colours of edges in some alternating path V0,V1,...,Vi, where i

Dear Aleksey Nikolaevich Glebov ,

      Your statement of  the problem is exactly right. As you may know, I am not major in Graph theory. I‘m still confused by some statements in your latest proof. I will send an email to you and explain the background about this question. Maybe we can solve this problem completely together if you are interested.

I am open to any questions and discussions. I will wait for a message from you.