Approximating min with p-norm For nonnegative $x \in \mathbb{R}_+^n$, $p \geq 1$ I want to find $\epsilon_{n,p}$ such that
$$
\frac{n^{1/p}}{||1/x||_p} \geq \min (x) \geq \frac{n^{1/p}}{||1/x||_p} - \epsilon_{p,n} ||x||_\infty
$$
Here is how far I got.  Without loss of generality, take $||x||_\infty=1$, furthermore, let $x$ be the vector $(m,1,...,1)$, where $m$ is the minimum element of $x$.  Then our goal is to find the maximum value of
$$\epsilon_{n,p}(m) = \left(\frac{n}{p - 1 + m^{-p}}\right)^{1/p} - m$$
What is an upper bound on $\epsilon_{n,p}(m)$ on $m \in [0,1]$?
See this plot for n=3, p=3
 A: [Edit to answer the modified question:]
For $p=1$, $\epsilon_{n,p}(m)=(n-1)m$ and there is no upper bound on $\epsilon$, so I will assume $p>1$.
The derivative of this new version of $\epsilon$ with respect to $m$ is
$$\epsilon'_{n,p}(m) = n^{1/p}m^{-(p+1)}\left(p - 1 + m^{-p}\right)^{-(p+1)/p} - 1\;.$$
Setting this to zero yields
$$n^{1/p}m^{-(p+1)}\left(p - 1 + m^{-p}\right)^{-(p+1)/p}=1\;,$$
$$n^{-1/(p+1)}m^{p}\left(p - 1 + m^{-p}\right)=1\;,\tag{1}$$
$$(p - 1)m^p + 1=n^{1/(p+1)}\;,$$
$$m=\left(\frac{n^{1/(p+1)}-1}{p-1}\right)^{1/p}\;.\tag{2}$$
Then using (1) to substitute $p - 1 + m^{-p}$ and (2) to substitute $m$ yields
$$
\begin{eqnarray}
\epsilon_{n,p}(m)
&=&
\left(n^{1/p}n^{-1/(p(p+1))}-1\right)m
\\
&=&
\left(n^{1/(p+1)}-1\right)\left(\frac{n^{1/(p+1)}-1}{p-1}\right)^{1/p}
\\
&=&
\left(\frac{\left(n^{1/(p+1)}-1\right)^{p+1}}{p-1}\right)^{1/p}\;.
\end{eqnarray}
$$
For $n=3$, $p=3$, this is about $0.171$, in agreement with your plot. To show that this is always a global maximum, note that $\epsilon_{n,p}(m)$ goes to $0$ for $m\to0$ and to $-\infty$ for $m\to\infty$, so if the derivative vanishes at a single point, it must be a global maximum.
