Relation between two p-norms While it's a well known that any two norms are equivalent for a finite dimensional normed linear space, I've been trying to derive the bounds for the case $X=\mathbb{R}^n$ and $l_p$-norms.
Let $1 \leq p,q <\infty.$ If $q>p$, from Holder's inequality we can derive that 
$$
\|x\|_p \leq (n)^{1/p-1/q}. \|x\|_q
$$
But I am unable to derive a similar $\leq$ tightest bound for the case $q<p$. Can anyone please provide a hint for this ?
 A: We have $ \lVert x \rVert_p  \leq \lVert x \rVert_1$ for all $1 \leq p < \infty $.
Why? Because $|x_1|^p + \ldots + |x_n|^p \leq (|x_1| + \ldots + |x_n|)^p$ (from Newton's binomial formula). 
So $(|x_1|^p + \ldots + |x_n|^p)^{1/p} \leq |x_1| + \ldots + |x_n|$.
For given $1 \leq q < \infty $ we have:
$|1 \cdot x_1| + \ldots + |x_n \cdot 1| \leq |1+ \ldots + 1|^{(1-1/q)} (|x_1|^q + \ldots + |x_n|^q)^{1/q} $
from Holder's formula.
Hence 
$(|x_1|^p + \ldots + |x_n|^p)^{1/p} \leq |x_1| + \ldots + |x_n| \leq n^{(1-1/q)} (|x_1|^q + \ldots + |x_n|^q)^{1/q} \implies \lVert x \rVert_p  \leq n^{(1-1/q)} \lVert x \rVert_q$
which is true for all 
$1 \leq p,q < \infty $.
If $p > q$, the tighest bound is $1$: $ \lVert x \rVert_p \leq \lVert x \rVert_q$
Why? we need:$ \lVert x \rVert_p \leq \lVert x \rVert_q$ by dividng both sides by  $\lVert x \rVert_q$ we derive:
$\lVert \frac{x}{\lVert x \rVert_q} \rVert_p \leq 1 $
$ (| \frac{x_1}{\lVert x \rVert_q}|^p + \ldots + |\frac{x_n}{\lVert x \rVert_q}|^p)^{1/p}  \leq 1 $
$ | \frac{x_1}{\lVert x \rVert_q}|^p + \ldots + |\frac{x_n}{\lVert x \rVert_q}|^p  \leq 1 $
$ | \frac{x_1}{\lVert x \rVert_q}|^p + \ldots + |\frac{x_n}{\lVert x \rVert_q}|^p  \leq 
 | \frac{x_1}{\lVert x \rVert_q}|^q + \ldots + |\frac{x_n}{\lVert x \rVert_q}|^q = 1$.
This boundary couldn't less than $1$ beacuse for $x=(1,0,\ldots,0)$ we have equality:
 $ \lVert x \rVert_p = \lVert x \rVert_q$.
