Bounding the $l_1$ norm of a vector

Let $x$ be real vector with $\|x\|_1=x_1+\ldots +x_{2n}$.

How to bound from above $(x_1+\ldots+x_n)(x_{n+1}+\ldots+x_{2n})$ by $l_2$ norm of the vector $x$.

Of course, using $\|x\|\leq\sqrt {2n}\|x\|_2$ I can bound $$(x_1+\ldots+x_n)(x_{n+1}+\ldots+x_{2n})\leq\|x\|^2_1\leq 2n\|x\|_2^2$$

But I would like to get an upper bount not greater then $1/2\|x\|_2^2$. Is it possible to get such a bound?

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If $x_j=1$, the LHS is $n^2$ and $1/2||x||^2_2=n/2$. Hence the constant you get is optimal in general. Why do you need such a bound? –  Davide Giraudo Aug 1 '12 at 21:51
Yes, but I would like to get bound for any vector $x$ –  user 321123 Aug 1 '12 at 21:54
Actually that should be $(n/2) \|x\|_2^2$. Note that if all $x_i = 1$, $\|x\|_2^2 = 2n$. –  Robert Israel Aug 1 '12 at 22:11
Are you sure $x_1+\ldots+x_{2n}$ is a norm? Shouln't be $|x_1|+\ldots+|x_{2n}|$? –  enzotib Aug 2 '12 at 4:24

$$(x_1 + \ldots + x_n)(x_{n+1} + \ldots x_{2n}) = \sum_{i=1}^n \sum_{j=n+1}^{2n} x_i x_j \le \sum_{i=1}^n \sum_{j=n+1}^{2n} \frac{x_i^2 + x_j^2}{2} = \frac{n}{2} \|x\|_2^2$$ and this is an equality when all $x_i$ are equal.