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Given are $n$ real numbers $x(1)$, $x(2)$, ..., $x(n)$. Some of them are positive, some may be negative. The total sum is positive. Prove the following statement:

There exists some index $i$ such that all the following $n$ sums are positive:

$$x(i)$$ $$x(i) + x(i+1)$$ $$x(i) + x(i+1) + x(i+2)$$ $$...$$ $$x(i) + x(i+1) + x(i+2) + ... + x(i+n-1)$$ Here "plus" and "minus" within the brackets are meant modulo $n$.

Chain Markov
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For $0\le i\le n$, let $s(i)=x(1)+\cdots +x(i)$ (so in particular $s(0)=0$ and $s(n)>0$). Let $$m=\min\{s(0),s(1),\ldots,s(n)\} $$ and let $\iota$ be maximal with $0\le\iota\le n$ and $s(\iota)=m$. As $s(0)=0<s(n)$, it is clear that $m\le 0$ and $\iota<n$. Thus for $i:=\iota+1$ we have $1\le i\le n$, as desired. Then for $1\le j<i$, we find $$x(i)+\cdots+x(n)+x(1)+\cdots + x(j)=s(n)-s(\iota)+s(j)\ge s(n)>0$$ and for $i\le j\le n$, we have $s(j)>m$ and hence $$x(i)+\cdots+ x(j)=s(j)-s(\iota)>0.$$

Hagen von Eitzen
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  • Hello Hagen. Thanks for your response, but i have a question! i understand that you supposed a maximal value ι where m is minimum, but i cannot understand "what is J?" what does it refer to? THanks for your time! –  May 18 '19 at 11:08