How to prove this norm inequality?
Although the question could belong to SE.math, mastering inequalities for $\ell_p$ norms (for $p\ge 1$) or quasinorms (for $0&lt;p&lt; 1$), and their norm ratios and powers, is quite important in  modern signal/image processing: starting from Cauchy-Bunyakovsky-Schwarz, to power means, Rogers-Hölder, Young, Minkowski, Dresher-Gini etc.)  There are scalar, additive and convolutive versions, and reverse forms, that are used in data restoration and deconvolution, especially when using signal sparsity.

If $a_n$ and $b_n$ are positive finite sequences of length $N$, $u,v,w&gt;0$, and $1/u+1/v\le 1/w$, the generalized Rogers-Hölder's inequality gives:

\begin{equation}
\left( \sum_{n=1}^{N} (a_n b_n)^w \right)^{1/w} \le \left( \sum_{n=1}^{N} (a_n )^u \right)^{1/u}
\left( \sum_{n=1}^{N} (b_n )^v \right)^{1/v}\,.
\label{eq_rogers-holder}
\end{equation}

You can find a proof in P. S. Bullen, Handbook of Means and Their Inequalities, Kluwer Academic Publishers, 2nd edition, 2003, p. 187, Corollary 8.
22, 

If you now set $a_n=|z_n|^p$, $b_n=1$, $w=p$, $v=q$ and  $1/u=1/p-1/q$, you get 
\begin{equation}
\ell_{q}{(|z|)} \leq \ell_{p}{(|z|)}  \leq \ell_{0}^{\frac{1}{p}-\frac{1}{q}}\ell_{q}{(|z|)}   \,.
\end{equation}

$\ell_{0}$ here, not a norm, nor a quasinorm, is the count index, ie the number of nonzero terms in $z$, or the sparsity. In other words, there is  an inclusion in $l_p$ spaces: if $p\le q$, $l_p\subset l_q$.
Your answer is incorrect, because you should have $a\le b$ instead. 

One of the basic block is alluded to by  @robert bristow-johnson. If $0&lt;c\le 1$ ($a_n$ positive as above):

$$ \left(\sum a_n\right)^c \le  \sum a_n^c \,.$$ 

Using additivity, prove it first with two terms:

$$(a_1+a_2)^c \le a_1^c+a_2^c\,.$$

By factoring by the largest of $a_1$ or $a_2$, you end up to prove that for $0&lt;t&lt;1$:

$$(1+t)^c\le 1+t^c\,.$$

Just differentiate the function $t\mapsto 1+t^c- (1+t)^c$ as $c\left(t^{c-1}- (1+t)^{c-1}\right)$, and verify the latter is positive.