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

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

$x(i)$

$x(i)+x(i+1)$

$x(i)+x(i+1)+x(i+2)$

$...$

$x(i)+x(i+1)+x(i+2)+...+x(i+n−1)$

Here "plus" and "minus" within the brackets are meant modulo $n$.

nmasanta
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Abd El-Rahman Ahmed
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1 Answers1

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One can prove it by induction.

For $n=1$ and $n=2$ the statement is clear. Now suppose $n>2$. We may assume that at least one number is positive, since otherwise the statement is clear. On the other hand at least one number must be negative since the total sum is negative. Therefore there is an index $j$ such that $x(j)$ is negative and $x(j+1)$ (+ modulo n) is positive. Now replace these two numbers by their sum $x(j)+x(j+1)$. For the thus reduced sequence of numbers we know that there is a solution $i$ by induction hypotheses. Clearly this is also a solutions for the original problem.

I hope I made no mistake.

Reinhard Stahn
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