TITLE: Interesting property of finite integer sequence with sum 1
QUESTION [5 upvotes]: I ran into nice problem which I proved and would be thankful if anyone assess it.
Given a sequence of integers $x_1,  x_2, \dots, x_n$ whose sum is 1? prove that exactly one of the cyclic shifts 
$$x_1,x_2,\dots, x_n; \quad x_2,\dots, x_n,x_1; \cdots  \quad x_n, x_1\dots, x_{n-1}$$ has all of his partial sums positive. (By a partial sum we mean the sum of the first $k$ terms, $k\leqslant n$.)
My solution: For $n=1$ we have a single-term sequence and it is obvious.
Assume that our proposition is true for $n$. We'll try to prove it for $n+1$.
Let we have the sequence $(x_1, x_2, \dots, x_n,x_{n+1})$ then $\exists \ i$ such that $x_i>0$.

If $1\leqslant i <n+1$ then consider the following sequence $(x_1, \dots,x_i+x_{i+1},\dots, x_{n+1})$ which has $n$ elements. Then using the assumption of induction we have that there is exactly one cyclic shift with desired property, And if we write $x_i, x_{i+1}$ instead of $x_i+x_{i+1}$ and since $x_i>0$ we get our desired sequence.
If $i=n+1$, i.e. $x_{n+1}>0$ then replace the sequence $(x_1, x_2, \dots, x_n,x_{n+1})$ by $(x_2, \dots, x_n,x_{n+1}+x_1)$ which consists of $n$ terms. Then using the assumption of induction we get exactly one sequence with desired properties and separating $x_{n+1}+x_1$ to $x_{n+1}, x_1$ we get the sequence with $n+1$ elements since $x_{n+1}>0$.

REPLY [3 votes]: In the above post I have shown the existence of cyclic shift of the sequence with positive partial sums. Right now I am goint to show that such sequence is unique.
However, my proof is not related with the induction proof.
Let's we have sequence $(x_1,x_2,\dots,x_n)$. One of its cyclic shift which has all of its partial sums positive WLOG let be $(x_1,x_2,\dots,x_n)$ and another is $(x_i,x_{i+1},\dots, x_{n+i-1})$ for some $i\in\{2,\dots,n\}$. Then $x_1+\dots +x_{i-1}>0$ i.e. $x_1+\dots +x_{i-1}\geqslant 1$ and also $x_i+\dots+x_n>0$ i.e. $x_i+\dots+x_n\geqslant 1$ then adding these two inequalities $1=x_1+\dots+x_n\geqslant 2$ which is contradiction. Thus we proved the uniqueness.