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?

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