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\left|z\_j\right|^p\right|\leq C\_{p,l}\sum\_{i\neq j}\left|z\_i\right|\left|z\_j\right|^{P-1}.
\end{equation}



A homogeneous but slightly asymmetric inequality
Here is a somewhat silly way to do it. You need to prove that:


$$\frac{\left|\sum\_{j=1}^l z\_j\right|^p-\sum\_{j=1}^l\left|z\_j\right|^p}{\sum\_{i\neq j}\left|z\_i\right|\left|z\_j\right|^{P-1}}$$


is bounded function on the unit sphere $S^{l-1}$. $S^{l-1}$ 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.


