# Partition of numbers into two subsets with some properties

I am trying to prove the following. Fix $$n\geq2$$ weakly positive numbers $$x_{1},\ldots,x_{n}$$ that sum to some constant $$c>0$$ and integer $$r\in\{1,\ldots,n-1\}$$. Suppose the numbers are weakly increasing (i.e., $$x_{i}\leq x_{i+1}$$ $$\forall i\in\{1,\ldots,n-1\}$$). Then, I am trying to prove, one can partition $$\{1,\ldots,n-1\}$$ into two subsets $$A$$ and $$B$$ such that $$\sum_{i\in A}x_{i}\leq\frac{1}{n-r+1}c$$ and $$\sum_{i\in B}x_{i}\leq\frac{n-r}{n-r+1}c$$.

Context: I am trying to prove this property as part of a research project I am working on. The nature of the project will not be informative for thinking about what I am trying to prove. However, the partition problem (with weakly positive numbers that are not integers), keeps appearing in what I am trying to do.

So far: A possible first step seems to be the following observation. Suppose that the largest of the original $$n$$ numbers, $$x_{n}$$, satisfies $$x_{n}\geq\frac{1}{n-r+1}c$$. Then the sum of the $$n-1$$ remaining numbers is weakly less than $$c-\frac{1}{n-r+1}c=\frac{n-r}{n-r+1}c$$, and hence the postulated partition is $$A=\emptyset$$ and $$B=\{1,\ldots,n-1\}$$. The problem is I am not sure how to proceed for the $$x_{n}<\frac{1}{n-r+1}c$$ case.

Any ideas, pointers to related problems, papers, etc. will be greatly appreciated.

Let $$s$$ be the largest positive integer less than $$n$$ such that $$\sum_{i=1}^sx_i\leq {c\over n-r+1}.$$ (Clearly $$1$$ is such an integer.) If $$s=n-1$$ we are done for we can take $$B=\emptyset.$$ I claim $$A =\{1,2,\dots,s\}, B=\{s+1,s+2,\dots,n-1\}$$ is an acceptable partition. Suppose not, that is, suppose $$\sum_{i=s+1}^{n-1}x_i>{n-r\over n-r+1}c\tag{1}$$ By definition of $$s$$ $$\sum_{i=1}^{s+1}x_i>{c\over n-r+1}\tag{2}$$ Adding $$(1)$$ and $$(2)$$ gives $$x_{s+1}+\sum_{i=1}^{n-1}x_i>c=x_n+\sum_{i=1}^{n-1}x_i\implies x_{s+1}>x_n$$ which is a contradiction.