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$\Bigl( \sum\limits_{i=1}^{n-1} a_{i}^{2}\Bigr) \cdot \Bigl( \sum\limits_{j=1}^{n-1} b_{i}^{2}\Bigr)-\Bigl( \sum\limits_{l=1}^{n-1} a_{l}b_{l}\Bigr)^{2}-2\cdot\lceil\frac{n-1}{2} \rceil\cdot \Bigl( \sum\limits_{l=1}^{n-1} a_{l}b_{l}\Bigr)-2\cdot\lfloor\frac{n-1}{2} \rfloor\cdot \Bigl( \sum\limits_{l=1}^{n-1} a_{l}b_{l}\Bigr)-\\ -\lfloor\frac{n-1}{2} \rfloor \cdot \lceil\frac{n-1}{2} \rceil \leq 0,$

where $|a_{l}-b_{l}|\leq 1 \hspace{3mm} \forall l=\overline{1, n-1}$ and $a_{l}, b_{l} =\{0,1,2,\ldots\}$.

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Welcome to math.SE. Since you are a new user, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. –  Antonio Vargas May 19 '12 at 20:39
    
It seems to me, that I've found the solution of the problem.] –  Sh.N. May 19 '12 at 21:08
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You can post your solution as an answer to this question, if you'd like. :) –  Antonio Vargas May 19 '12 at 21:53
    
@Antonio Vargas Of course :) –  Sh.N. May 19 '12 at 22:35
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