# the existence of this number $(a_{1}+a_{2}\ldots +a_{n})^{2}-(n^2-n+2)a_{i}a_{j} \geq 0$

Let $a_{1}, a_{2}, \ldots, a_{n}$, $n \geq 3$. Prove that at least one of the number $(a_{1}+a_{2}\ldots +a_{n})^{2}-(n^2-n+2)a_{i}a_{j}$ is greater or equal with $0$ for $1 \leq i < j \leq n$.

I don't know at least how to catch this problem . Thanks :)

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suppose it's not true, add everything together.

$t = \frac{n(n-1)}{2}$

$A = \sum a_i^2$

$B = \sum a_ia_{i+1}$

$tA < 2B$

Which is a contradiction because $t > 2$ and $A \geq B$.

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from where $t$? – Iuli Jan 11 '13 at 23:01
I think I know, let me try. thanks – Iuli Jan 11 '13 at 23:03
t is the C(2, n), the number of equations – Bojan Serafimov Jan 11 '13 at 23:04

Suppose for contradiction that the condition didn't hold. This gives you a set of ${n \choose 2}$ sharp inequalities in the $(i,j)$ which you can multiply together, to yield a sharp inequality of homogeneous polynomials of degree $n(n-1)$. However, the reverse inequality can be proven via "bashing" techniques, so it shows that your hypothesis was false - meaning that the condition holds for some $(i,j)$ pair, which is what we were trying to show.

The reason I am being vague is because I don't think this is your own problem, and you haven't given an attribution - so I am unsure if it is an open contest problem, for instance.

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