Why this power inequality for sums of real numbers holds? $$\left|\sum_{i=1}^nx_i\right|^p \leq \begin{cases} \sum_{i=1}^n|x_i|^p & p\in(0,1]\\ n^{p-1}\sum_{i=1}^n|x_i|^p & p>1 \end{cases}$$
Can it be generalized for arbitrary sequences $\{x_i\}_{i=1}^n$ in Hilbert spaces in case $p=2$?
 A: The first inequality can be shown with induction if we show
$$(1+z)^p \leq 1+z^p$$
for $z>1$ and then let $z=x/y$.
A: *

*For $p > 1$, apply Holder's Inequality to 
$$\left\lvert \sum_{k=1}^n x_i\right\rvert \leq \sum_{k=1}^n 1\cdot \lvert x_i\rvert \leq n^{1/q}  \left(\sum_{k=1}^n \lvert x_i\rvert^p\right)^{1/p} $$
with $q =\frac{p}{p-1}$ is the Holder conjugate of $p$. Now, raise both sides to the power $p$ to get the inequality.

*For $p \in (0,1]$, this comes from the following fact:

Lemma. For any sequence $x=(x_1,\dots,x_n)\in\mathbb{R}^n$, $p > 0 \mapsto \lVert{x}\rVert_p$ is non-increasing. In particular, for $0 < p \leq q <\infty$,
$$
\left(\sum_i |{x_i}|^q\right)^{1/q} = \lVert{x}\rVert_q \leq \lVert{x}\rVert_p = \left(\sum_i |{x_i}|^p\right)^{1/p}\;.
$$ To see why, one can easily prove that if $\lVert{x}\rVert_p = 1$, then $\lVert{x}\rVert_q^q \leq 1$ (bounding each term $|{x_i}|^q \leq |{x_i}|^p$), and therefore $\lVert{x}\rVert_q \leq 1 = \lVert{x}\rVert_p$. Next, for the general case, apply this to $y = x/\lVert{x}\rVert_p$, which has unit $\ell_p$ norm, and conclude by homogeneity of the norm.

By an application of the triangle inequality and the above, taking $q=1$ you get 
$$
\left\lvert\sum_i {x_i}\right\rvert \leq \sum_i \lvert{x_i}\rvert = \lVert{x}\rVert_1 \leq \lVert{x}\rVert_p = \left(\sum_i |{x_i}|^p\right)^{1/p}
$$
and it suffices once again to raise both LHS and RHS to the $p$-th power to conclude.
