How do you show monotonicity of the $\ell^p$ norms? I can't seem to work out the inequality $(\sum |x_n|^q)^{1/q} \leq (\sum |x_n|^p)^{1/p}$ for $p \leq q$ (which I'm assuming is the way to go about it).
 A: You are right @user1736.
If $0<a\leq1$ then 
$$\left(\sum |a_n|\right)^a\leq \sum|a_n|^a.\tag{1}$$
Hence for $p\leq q$ we have $p/q\leq1$, and
$$\left(\sum_n |x_n|^q\right)^{1/q}=\left(\sum_n |x_n|^q\right)^{p/qp}\leq \left(\sum_n |x_n|^{q(p/q)}\right)^{1/p}=\left(\sum|x_n|^p\right)^{1/p}$$

Edit:  How do we prove (1) (for $0<a<1$)?
Step 1. It is sufficient to prove this for finite sequences because then we may take limits.
Step 2. To prove the statement for finite sequences it is sufficient to prove 
$$(x+y)^a\leq x^a+ y^a,\quad\text{ for $x,y>0$}\tag{2}$$
because the finite case is just iterations of (2). 
Step 3. To prove (2) it suffice to prove $$(1+t)^a\leq 1+t^a,\quad\text{ where $0<t<1$} \tag{3}$$
Now, the derivative of the function $f(t)=1+t^a-(1+t)^a$ is given by $f'(t) = a(t^{a-1} - (1+t)^{a-1})$ and it is positive since $a>0$ and $t\mapsto t^b$ is decreasing for negative $b$. Hence, $f(t)\geq f(0)=0$ for $0<t<1$, which proves (3).
A: The $\boldsymbol{\ell^{2^m}}$ Norm is Less Than the $\boldsymbol{\ell^1}$ Norm
Assume $b_k\ge0$,
$$
\begin{align}
\left(\sum_{k=1}^nb_k\right)^2
&=\sum_{k=1}^nb_k^2+2\!\!\!\sum_{\substack{j,k=1\\j\lt k}}^n\!\!b_jb_k\\
&\ge\sum_{k=1}^nb_k^2\tag1
\end{align}
$$
Therefore, by induction, we have that
$$
\left(\sum_{k=1}^nb_k\right)^{2^m}\ge\sum_{k=1}^nb_k^{2^m}\tag2
$$

Interpolate
For $1\lt r\lt2^m$,
$$
\begin{align}
\sum_{k=1}^nb_k^r
&=\sum_{k=1}^n\left(b_k^{2^m}\right)^{\frac{r-1}{2^m-1}}b_k^{\frac{2^m-r}{2^m-1}}\tag3\\
&\le\left(\sum_{k=1}^nb_k^{2^m}\right)^{\frac{r-1}{2^m-1}}\left(\sum_{k=1}^nb_k\right)^{\frac{2^m-r}{2^m-1}}\tag4\\
&\le\left(\sum_{k=1}^nb_k\right)^{2^m\frac{r-1}{2^m-1}}\left(\sum_{k=1}^nb_k\right)^{\frac{2^m-r}{2^m-1}}\tag5\\
&=\left(\sum_{k=1}^nb_k\right)^r\tag6
\end{align}
$$
Explanation:
$(3)$: $r=2^m\frac{r-1}{2^m-1}+\frac{2^m-r}{2^m-1}$
$(4)$: Hölder
$(5)$: Inequality $(2)$
$(6)$: $r=2^m\frac{r-1}{2^m-1}+\frac{2^m-r}{2^m-1}$

Apply to $\boldsymbol{\ell^p}$ and $\boldsymbol{\ell^q}$
Let $r=\frac qp$, and $b_k=a_k^p$, then $(6)$ says
$$
\sum_{k=1}^na_k^q\le\left(\sum_{k=1}^na_k^p\right)^{q/p}\tag7
$$
which is equivalent to
$$
\left(\sum_{k=1}^na_k^q\right)^{1/q}\le\left(\sum_{k=1}^na_k^p\right)^{1/p}\tag8
$$
A: Let $|x_i|^p=a_i$ and $\frac{q}{p}=k$. 
Thus, $k\geq1$ and we need to prove that:
$$(a_1+a_2+...+a_n)^k\geq a_1^k+a_2^k+...+a_n^k.$$
Now, let $a_1\geq a_2\geq...\geq a_n$.
Thus, since $f(x)=x^k$ is a convex function and $$(a_1+a_2+...+a_n,0,...,0)\succ(a_1,a_2,...,a_n),$$ by Karamata we obtain:
$$(a_1+a_2+...+a_n)^k+0^k+...+0^k\geq a_1^k+a_2^k+...+a_n^k,$$ which is our inequality.
A: I don't think you need to prove the inequality you have in the question; that's a bit too strong. Note that $\{x_n\}\in\ell_p$ if and only if $\left(\sum|x_n|^p\right)^{1/p}$ is finite, if and only if $\sum|x_n|^p\lt\infty$. So you really just need to show that if $\sum|x_n|^p$ is finite, then $\sum|x_n|^q$ is finite, assuming $p\leq q$. 
You want to remember is two things:


*

*if $p\leq q$, then for $|x|>1$ you have $|x|^p\leq|x|^q$, but if $|x|<1$, then $|x|^p \geq |x|^q$.

*$\sum_{n=1}^{\infty}a_n$ converges if and only if for every $m\geq 1$, $\sum_{n=m}^{\infty}a_n$ converges.

A: For completeness I will add this as an answer (it is a slight adaptation of the argument from AD.):
For $a\in[0,1]$ and any $y_i\geq 0, i\in\mathbb N$, with at least one $y_i\neq0$ and the convention that $y^0=1$ for any $y\geq0$, \begin{equation}\label{*}\tag{*}\sum_{i=1}^\infty \frac{y_i^a}{\left(\sum_{j=1}^\infty y_j\right)^a}=\sum_{i=1}^\infty \left(\frac{y_i}{\sum_{j=1}^\infty y_j}\right)^a\geq \sum_{i=1}^\infty \frac{y_i}{\sum_{j=1}^\infty y_j}=1,\end{equation}
where I have used $y^a\geq y$ whenever $y\in[0,1]$ and $a\in[0,1]$. (This can be derived for instance from the concavity of $y\mapsto y^a$.)
For $p=q$, there is nothing to prove. For $1\le p< q\le\infty$ and $x=(x_i)_{i\in\mathbb N}\in \ell^q$, set $a\overset{\text{Def.}}=\frac pq\in[0,1]$ and $y_i\overset{\text{Def.}}=\lvert x_i\rvert^q\ge0$. Then \eqref{*} yields
\begin{equation*}
\sum_{i=1}^\infty \lvert x_i\rvert^p\geq\left(\sum_{i=1}^\infty \lvert x_i\rvert^{q}\right)^{\frac pq},
\end{equation*}
i.e.
\begin{equation*}
\lVert x\rVert_{\ell^q}\le\lVert x\rVert_{\ell^p}.
\end{equation*}
