Something from a $\LaTeX$ file I have somewhere, posting it here as "Community Wiki" answer for further use, in case it proves helpful.
Theorem. For any sequence $x=(x_1,\dots,x_n)\in\mathbb{R}^n$, $p > 0 \mapsto \lVert x\Vert_p$ is non-increasing. In particular, for $0 < p \leq q <\infty$,
$$
\left(\sum_i \lvert{x_i}\rvert^q\right)^{1/q} = \lVert x\Vert_q \leq \lVert x\Vert_p = \left(\sum_i \lvert{x_i}\rvert^p\right)^{1/p}\;.
$$
Proof (sketch). To see why, one can easily prove that if $\lVert x\Vert_p = 1$, then $\lVert x\Vert_q^q \leq 1$ (bounding each term $\lvert{x_i}\rvert^q \leq \lvert{x_i}\rvert^p$), and therefore $\lVert x\Vert_q \leq 1 = \lVert x\Vert_p$. Next, for the general case, apply this to $y \stackrel{\rm def}{=} \frac{x}{\lVert x\Vert_p}$, which has unit $\ell_p$ norm, and conclude by homogeneity of the norm.
(The above deals with $x\neq 0$. Note that the case $x=0$ is immediate.)
Finally, the case $q=\infty$ is straightforward, as for any $p < \infty$ and any $x\in\mathbb{R}^n$ we have $\lVert x\rVert_\infty = \max_i \lvert{x_i}\rvert = (\max_i \lvert{x_i}\rvert^p)^{1/p} \leq \left(\sum_{i} \lvert{x_i}\rvert^p\right)^{1/p} = \lVert x\rVert_p.$