Proving an inequality with $\|x\|_p$ metrics? Let $1 \leq p < q \leq \infty$ and $x \in\mathbb{R}^n$. Show that $\|x\|_q \leq \|x\|_p \leq n^\frac{1}{p}\|x\|_q$, where $\|x\|_p$ is the metric $\left(\sum_{j=1}^n{|x_j|^p}\right)^\frac{1}{p}$. 
A hint is given: "For the left-hand inequality do first the case where $\|x\|_p = 1$, and for the right-hand inequality do first the case $\|x\|_q = 1$." So first I set $\|x\|_p = 1$ and got $\sum_{k = 1}^n|x_k|^q \leq 1 \leq n^\frac{q}{p}\sum_{k = 1}^n|x_k|^q$. This makes sense since $\frac{q}{p} > 1$. 
I left this and used the other half of the hint; I considered the inequality again and set $\|x\|_q = 1$ to obtain $1 \leq \sum_{k = 1}^n|x_k|^p \leq n$. I'm not quite as sure what this inequality means.
My question is, how to extend these two inequalities to the more general cases where $\|x\|_p, \|x\|_q \neq 1$? And how can I make my two results relate to each other, since they seem to be two completely different cases? Perhaps this question is simpler than I am making it out to be and I might not be wording myself clearly but once I understand the general principle behind inequalities like this I will be able to do more complex ones on my own.
 A: Note: The hint is that you should consider the special case $\lVert x\rVert_p = 1$ in order to prove that in that case, you have
$$\lVert x\rVert_q \leq \lVert x\rVert_p,$$
ignoring the second inequality; and then, separately, that you consider the case $\lVert x\rVert_q=1$ to prove that in that case
$$\lVert x\rVert_p \leq n^{1/p}\lVert x\rVert_q,$$
while ignoring the first inequality

First: How do you prove these special cases?
Assume first that $\lVert x\rVert_p = 1$, so that
$$\sum_{j=1}^n|x_i|^p = 1.$$
That means that $0\leq |x_i|\leq 1$ for all $i$, and since $q\gt p$, then
$$0\leq |x_i|\leq 1\Rightarrow 0\leq |x_i|^q \leq |x_i|^p \leq 1.$$
Hence $\sum |x_i|^q \leq \sum |x_i|^p =1$, and taking $q$th roots you conclude that $\lVert x \rVert_q \leq 1 = \lVert x\rVert_p$. 
Try something along those lines for the second inequality, starting from the assumption that $\lVert x \rVert_q =1$.

Second: how do you extend the special case to the general case?
Given an arbitrary $x\neq \mathbf{0}$, let $\lambda=\frac{1}{\lVert x\rVert_p}$. If you already know the inequality when the $p$-norm is $1$, then you know that
$$\lVert \lambda x \rVert_q \leq \lVert \lambda x \rVert_p,$$
since the $p$-norm of $\lambda x$ is $1$. But since $\lambda$ is a positive scalar, this is equivalent to
$$\lambda\lVert x \rVert _q \leq \lambda \lVert x\rVert_p,$$
and cancelling $\lambda$ gives the desired inequality for arbitrary $x\neq\mathbf{0}$. 
A similar trick works for the second inequality. And of course, the inequality trivially holds if $x=\mathbf{0}$. 
