Derive $\frac1n \|x\|_p^p \leq \|x\| \leq n^{p/2}\|x\|_p^p$ from Holder's inequality? Given a vector $x = (x_1, \dotsc, x_n)\in \mathbb{C}^n$, I wanted to compare $|x_1|^p + \dotsb + |x_n|^p$ to $\|x\|^p$. I discovered that if $m=\max_i|x_i|$, we have 
$$m^p \leq \|x\|^p \leq n^{p/2}m^p$$
and
$$m^p \leq |x_1|^p + \dotsb + |x_n|^p \leq nm^p,$$
which gives us 
$$\frac1n (|x_1|^p + \dotsb + |x_n|^p) \leq \|x\|^p \leq n^{p/2}(|x_1|^p + \dotsb + |x_n|^p).$$
Then I noticed I could just express this as
$$\frac1n \|x\|_p^p \leq \|x\|^p \leq n^{p/2}\|x\|_p^p.$$
Could I have derived this easily from Holder's inequality in a way I'm missing?
 A: There are really four inequalities here, two of Hölder type and two non-Hölder-type inequalities.  I say that an inequality is non-Hölder if it depends on the structure of the measure space in question; this meaning should be clear from the discussion below. Our measure space is the set $\{1,\dots,n\}$ with the counting measure. 
When $p\ge 2$, the inequality $\|x\|^p\le n^{p/2} \|x\|_p^p$ is of Hölder type. Indeed, 
$$\sum x_i^2\cdot 1 \le (\sum |x_i|^{p})^{2/p} \left(\sum 1\right)^{1-2/p} \tag1$$ 
which after raising to power $p/2$ yields 
$$\|x\|^p\le n^{p/2-1}\|x\|_p^p,\quad p\ge 2 \tag2$$
Observe that the constant in (2) is better (smaller) than $n^{p/2}$ in your question. 
When $1\le p\le 2$, the inequality $\|x\|^p\ge n^{-1} \|x\|_p^p$ is also of Hölder type. Indeed, 
$$\sum |x_i|^p\cdot 1 \le (\sum |x_i|^2)^{p/2} \left(\sum 1\right)^{1-p/2} \tag3$$ 
which can be written as 
$$n^{p/2-1}\|x\|_p^p\le  \|x\|^p,\quad 1\le p\le 2 \tag4$$
Observe that the constant in (4) is  better (larger) than $n^{-1}$ in your question. 
Being derived from Hölder's inequality, estimates (2) and (4) are independent of the structure of the measure space: they apply equally well if $x$ is a measurable function on some measure space $(X,\mu)$ with $\mu(X)=n$. 
The other two inequalities 
$$n^{ -1}\|x\|_p^p\le  \|x\|^p,\quad p\ge 2  \tag5$$
and 
$$\|x\|^p\le n^{p/2 }\|x\|_p^p,\quad 1\le p\le 2\tag6$$
do not hold for functions on an arbitrary measure space $(X,\mu)$ with $\mu(X)=n$. (You can build counterexamples on the interval $[0,n]$ in the form of $x(t)=t^r$ for appropriate $r<0$.) These rely on the discrete nature of the particular space $\{1,\dots,n\}$. I think this qualifies them as non-Hölder-type  inequalities.
You may also observe the difference of extremizers: in (2) and (4) they are   vectors with components of equal magnitude; in (5) and (6) they are standard basis vectors.
