Generalize the binomial equation I am wondering if generalizing the case $p=2$ we have for $1<p<\infty$ that there is a constant $c(p,l)$ such that for all $(z_j)$ with $z_j \in \mathbb C$ that
$$\left\lvert \left\lvert \sum_{j=1}^l z_j \right\rvert^p - \sum_{j=1}^l \left\lvert z_j \right\rvert^p \right\rvert \le c(p,l) \sum_{j \neq k} \vert z_j \vert \vert z_k \vert^{p-1}$$
in the case $p=2$ I can clearly see it holds, but here, I don't know.
 A: Suppose that not all $z_i$ are zero. Without loss of generality assume that $z_1$ has the largest absolute value, and by homogeneity, assume that $|z_1| = 1$. Let $\delta = \sum_{j \geq 2} |z_j|$, so $\delta \leq l-1$, and note that the RHS of the desired inequality is at least $c(p, l)\delta|z_1|^{p-1} = c(p, l)\delta$.
Let's find bounds on the two sums from the LHS in terms of $\delta$. For the first sum, by the triangle inequality we always have $0 \leq \left|\sum z_j\right|^p \leq (1 + \delta)^p$, and when $\delta \leq 1$, we have
$(1 - \delta)^p \leq \left|\sum z_j\right|^p$ as well. Now, for any $\delta$ we have
$$(1 + \delta)^p - 1 = \int_0^\delta p(1 + x)^{p-1} \,dx \leq \delta p(1 + \delta)^{p-1},$$
and when $\delta \leq 1$,
$$1 - (1 - \delta)^p = \int_0^\delta p(1 - x)^{p-1} \,dx \leq \delta p \leq \delta p (1 + \delta)^{p-1}$$
as well, hence $|\left|\sum z_j\right|^p - 1| \leq \delta p(1 + \delta)^{p-1}$ in every case.
The other sum satisfies $1 \leq \sum |z_j|^p \leq \sum |z_j| \leq 1 + \delta$, hence $\left| \sum |z_j|^p - 1\right| \leq \delta$, so it follows that the LHS of our desired inequality is bounded by $\delta p(1 + \delta)^{p-1} + \delta \leq pl^{p-1}\delta + \delta$. Since our RHS was at least $c(p, l)\delta$, it suffices to take $c(p, l) = pl^{p-1} + 1$.
This choice of $c(p, l)$ is probably far from optimal. In the case where all $z_j = 1$, we see that $c(p, l)$ must be at least $\approx l^{p-2}$, and I would guess this is closer to optimal.
