I need to prove the following inequality: for any $Z=(z_1,\dots,z_l)\in\mathbb{C}^l$ for any $p\geq 2$ and $l\geq 2$ \begin{equation} \left\left\sum_{j=1}^l z_j\right^p\sum_{j=1}^l\leftz_j\right^p\right\leq C_{p,l}\sum_{i\neq j}\leftz_i\right\leftz_j\right^{P1}. \end{equation}

$\begingroup$ Why do you "need" to do anything?  Is this homework? $\endgroup$– Stefan Kohl ♦Jan 19, 2014 at 18:59

$\begingroup$ No it is not homework. I was trying to understand a paper of Cazenave were he proved a scattering property for focusing nonlinear Schrödinger equation. I found this inequality not trivial. $\endgroup$– Felice IandoliJan 19, 2014 at 19:40
1 Answer
Here is a somewhat silly way to do it. You need to prove that:
$$\frac{\left\sum_{j=1}^l z_j\right^p\sum_{j=1}^l\leftz_j\right^p}{\sum_{i\neq j}\leftz_i\right\leftz_j\right^{P1}}$$
is bounded function on the unit sphere $S^{l1}$. $S^{l1}$ is compact, and this function is continuous away from the places where $z_i=\pm 1$, where the denominator is $0$. So it is sufficient to prove boundedness in a neighborhood of those points. Say $z_1$ is very close to $1$. Then the denominator is very close to $p\sum_{j=2}^l z_j$ and the numerator is very close to $\sum_{j=2}^l z_j$. So the ratio is bounded.