A homogeneous but slightly asymmetric inequality involving $L_1,L_{p-1}$ and $L_p$ norms. I need to prove or disprove the following inequality:
for any $Z=(z_1,\ldots,z_l)\in\mathbb{C}^L$ and for any $p\geq 2$,
$$\biggl|\biggl\|\sum_{j=1}^L z_j\biggr\|^p - \sum_{j=1}^{L}\|z_j\|^p\biggr|\leq\frac{L^{p-1}-1}{L-1}\sum_{i\neq j}\|z_i\|\cdot\|z_j\|^{p-1}.\tag{1}$$
Here the term
$$\sum_{i\neq j}\|z_i\|\cdot\|z_j\|^{p-1} = \sum_{j=1}^{L}\|z_j\|\sum_{j=1}^{L}\|z_j\|^{p-1}-\sum_{j=1}^{L}\|z_j\|^p = \|Z\|_1\cdot\|Z\|_{p-1}^{p-1}-\|Z\|_{p}^{p}$$
is treatable, but the LHS in $(1)$ is not invariant if we replace, for istance, $z_k$ with $e^{i\theta_k}z_k$.
Nonetheless, since $(1)$ is homogeneous we can assume $\|\sum_{j=1}^{L}z_j\|=L$ or $\sum_{j=1}^L \|z_j\|=L$ without loss of generality.
For my purposes a weaker form of $(1)$, id est
$$\exists C_{p,L}:\forall Z\in\mathbb{C}^L,\quad\biggl|\biggl\|\sum_{j=1}^L z_j\biggr\|^p - \|Z\|_p^p \biggr|\leq C_{p,L}\cdot\sum_{i\neq j}\|z_i\|\cdot\|z_j\|^{p-1}\tag{2}$$
is enough, and I believe that $(2)$ follows, by some way, from Hanner's inequalities, but I did not manage to prove it. Writing
$$ z_k = A_k \exp(i\theta_k),\quad A_k\in\mathbb{R}^+ \tag{3}$$
may be useful in order to treat $(1)$ as an inequality between weigthed exponential sums.

When $p=2$, we have to prove:
$$ \left\|\sum z_i\right\|^2-\sum\|z_i\|^2 \leq \left(\sum \|z_i\|\right)^2-\sum\|z_i\|^2,\tag{4}$$
$$ \sum\|z_i\|^2-\left\|\sum z_i\right\|^2\leq \left(\sum \|z_i\|\right)^2-\sum\|z_i\|^2,\tag{5}$$
where $(4)$ follows from the triangle inequality and $(5)$ is equivalent, through $(3)$, to the trivial:
$$0\leq \sum_{i<j}2A_i A_j\left(1+\cos(\theta_i-\theta_j)\right).\tag{6}$$

In order to prove
$$\exists C_{p,L}:\forall Z\in\mathbb{C}^L,\quad\biggl\|\sum_{j=1}^L z_j\biggr\|^p - \|Z\|_p^p \leq C_{p,L}\cdot\sum_{i\neq j}\|z_i\|\cdot\|z_j\|^{p-1}\tag{2+},$$
assuming $\|z_1\|=1-\varepsilon$ and $\|Z\|_1=1$ we have to find $C_{p,L}$ such that:
\begin{equation*}1+(C_{p,L}-1)\|Z\|_p^p\leq C_{p,L}\|Z\|_{p-1}^{p-1},\tag{E1}\end{equation*}
so we only need to find a $C_{p,L}$ such that, setting  $W=(\|z_2\|,\ldots,\|z_L\|)$:
$$ C_{p,L}\geq \frac{1-(1-\varepsilon)^p-\|W\|_p^p}{\varepsilon(1-\varepsilon)^{p-1}+\|W\|_{p-1}^{p-1}-\|W\|_p^p} $$
$$ C_{p,L}\geq \frac{1-(1-\varepsilon)^p-\|W\|_p^p}{\varepsilon(1-\varepsilon)^{p-1}+(1-\varepsilon)\|W\|_{p-1}^{p-1}+\varepsilon\|W\|_{p-1}^{p-1}-\|W\|_p^p} $$
where $\|W\|_1=\varepsilon$. Since
$$\frac{1-(1-\varepsilon)^p-\|W\|_p^p}{\varepsilon(1-\varepsilon)^{p-1}+(1-\varepsilon)\|W\|_{p-1}^{p-1}+\varepsilon\|W\|_{p-1}^{p-1}-\|W\|_p^p} \leq \frac{1-(1-\varepsilon)^p-\|W\|_p^p}{\varepsilon(1-\varepsilon)^{p-1}+(1-\varepsilon)\|W\|_{p-1}^{p-1}+\frac{1}{C_{p,L-1}}\left(\varepsilon^p-\|W\|_p^p\right)} $$
holds by the induction hypothesis, it is enough to find $C_{p,L}$ such that:
$$C_{p,L}\geq \frac{1-(1-\varepsilon)^p-\|W\|_p^p}{\varepsilon(1-\varepsilon)^{p-1}+(1-\varepsilon)\|W\|_{p-1}^{p-1}}$$
or
$$C_{p,L}\geq \frac{1}{1-\varepsilon}\cdot\frac{\frac{1-(1-\varepsilon)^p}{\varepsilon}-\left(\frac{\varepsilon}{L-1}\right)^{p-1}}{(1-\varepsilon)^{p-2}+\left(\frac{\varepsilon}{L-1}\right)^{p-2}}.$$
Since the RHS is an increasing function with respect to $\varepsilon$ and $\varepsilon\leq\frac{L-1}{L}$, the RHS is bounded by its value in $\varepsilon=\frac{L-1}{L}$, so the choice
$$C_{p,L}=\frac{L}{2}\cdot\frac{L^{p-1}-1}{L-1}$$
is effective. This constant is just $L/2$ times the optimal constant.
 A: Here there is a roadmap to prove $(1)$.
Two preliminary lemmas are needed:
$$\| Z \|_1^p \leq L^{p-1}\|Z\|_p^p\,, \tag{A}$$
$$ \frac{1}{L}\|Z\|_1 \|Z\|_{p-1}^{p-1} \leq \|Z\|_p^p \leq \|Z\|_1 \|Z\|_{p-1}^{p-1}.\tag{B}$$
$(A)$ follows from the Holder inequality, the right hand side of $(B)$ is trivial and the left hand side of $(B)$ follow from the logarithmic convexity of
$$ g(h) = \|Z\|_{h}^{h}.$$
In order to prove that $g$ is a log-convex function, $$g(h)g''(h)\geq g'(h)^2 \tag{C}$$ is sufficient, but $(C)$ is equivalent to:
$$\left(\sum\|z_k\|^h\right)\cdot\left( \sum\|z_k\|^h\log^2(\|z_k\|)\right) \geq \left(\sum\|z_k\|^h\log\|z_k\|\right)^2$$
that follows from the Cauchy-Schwarz inequality. Now there are two parts.

First part.
$$\left\|\sum z_k\right\|^p-\|Z\|_p^p\leq \frac{L^{p-1}-1}{L-1}\left(\|Z\|_1\|Z\|_{p-1}^{p-1}-\|Z\|_p^p\right)\tag{D}$$
follows from
$$\|Z\|_1^p-\|Z\|_p^p\leq \frac{L^{p-1}-1}{L-1}\left(\|Z\|_1\|Z\|_{p-1}^{p-1}-\|Z\|_p^p\right)$$
that is equivalent to:
$$(L-1)\|Z\|_1^p+(L^{p-1}-L)\|Z\|_p^p\leq (L^{p-1}-1)\|Z\|_1\|Z\|_{p-1}^{p-1}\tag{E}$$
or to:
$$L\|Z\|_1^p+L^{p-1}\|Z\|_p^p+\|Z\|_1\|Z\|_{p-1}^{p-1}\leq L^{p-1}\|Z\|_1\|Z\|_{p-1}^{p-1}+\|Z\|_1^p+L\|Z\|_p^p\tag{F}.$$
Notice that the first and last terms of the LHS and RHS satisfy the inequality in the right direction ($\leq$), middle terms in the wrong one ($\geq$). Now I strongly believe that $(F)$, that resembles the Schur inequality, can be proved through Karamata's inequality. In terms of the $g$-function, $(F)$ becomes:
$$g(0)g(1)^p+g(0)^{p-1}g(p)+g(1)g(p-1)\leq g(0)^{p-1}g(1)g(p-1)+g(1)^p+g(0)g(p).\tag{G}$$
Writing $(E)$ in terms of the $A_k$s and assuming wlog $\sum A_k=L$, we can solve $(E)$ through Lagrange multipliers. The Lagrange conditions are
$$\lambda = A_k^{p-2}\left((L^{p-1}-L)pA_k-(L^{p-1}-1)(p-1)\right),$$
that sum up to:
$$L\lambda = p(L^{p-1}-L)\sum A_k^{p-1}-(L^{p-1}-1)(p-1)\sum A_k^{p-2}.$$
Unluckyly, the function 
$$l(x) = x^{p-2}\left((L^{p-1}-L)px-(L^{p-1}-1)(p-1)\right)$$
is not injective, so we cannot establish that the only stationary point of $(E)$ is $A_1=\ldots=A_L=1$. It is interesting to notice that, if we manage to prove that the function
$$t(p) = \left(\frac{\|Z\|_p}{\|Z\|_1}\right)^p+\left(\frac{\|Z\|_{p-1}}{L\,\|Z\|_1}\right)^{p-1}-\frac{1}{L^{p-1}}$$
or
$$r(p) = L^{p-1}\|Z\|_p^p + \|Z\|_1\|Z\|_{p-1}^{p-1}-\|Z\|_1^p \tag{H}$$
is decreasing, $(E)$ follows, too.

Second part.
We need to prove:
$$\|Z\|_p^p-\left\|\sum z_k\right\|^p\leq K\,\left(\|Z\|_1\|Z\|_{p-1}^{p-1}-\|Z\|_p^p\right)$$
that follows from
$$\|Z\|_p^p\leq\frac{K}{K+1}\|Z\|_1\|Z\|_{p-1}^{p-1},\tag{I}$$
but if we normalize the variables by setting $\|Z\|_1=\frac{K+1}{K}$, $(I)$ is trivial, so, in order to prove $(2)$, it is sufficient to prove $(2)$ without the absolute values in the left hand side, or to prove:
$$1-\sum_{k=1}^{L} A_k^p \leq C(p)\cdot\sum_{i\neq j}A_i A_j^{p-1}\tag{7}$$
under the hypothesis that the $A_k$s are non-negative numbers summing to $1$.

A reasonable trick is to differentiate $(7)$ with respect to $p$ and to consider that $C(2)=1$ works, so $(7)$ holds if
$$C(p)\sum_{i\neq j}-A_j^{p-1}A_i \log A_j+\sum_{k}-A_k^p \log A_k\leq C'(p)\sum_{i\neq j}A_i A_j^{p-1}\tag{8}$$
holds. Since $-x\log x$ takes values in $[0,1/e]$ over $[0,1]$, 
$$\frac{1}{e}\left(C(p)\sum_{i\neq j}A_j^{p-2}A_i +\sum_{k}A_k^{p-1}\right)\leq C'(p)\sum_{i\neq j}A_i A_j^{p-1}\tag{9}$$
is enough to prove $(8)$. We can write $(9)$ as:
$$C(p)\left(\|Z\|_{p-2}^{p-2}-\|Z\|_{p-1}^{p-1}\right) +\|Z\|_{p-1}^{p-1}\leq e\cdot C'(p)\,\left(\|Z\|_{p-1}^{p-1}-\|Z\|_{p}^{p}\right),\tag{10}$$
or:
$$C(p)+\frac{g(p-1)}{g(p-2)-g(p-1)}\leq e\cdot C'(p)\cdot\frac{g(p-1)-g(p)}{g(p-2)-g(p-1)}.\tag{11}$$
