Do $\ell_p$ distances contract when scaling down onto unit $\ell_q$ ball? More specifically, is it true that
$\Vert\frac{x}{\max\{1, \|x\|_q\}}-\frac{y}{\max\{1, \|y\|_q\}}\Vert_p\leq\Vert x-y\Vert_p$ for all $x, y$ when $p\geq 2$? 
($q$ is the obvious norm parameter satisfying $1/p+1/q=1$)
This is known to be true when $p=2$.
 A: Counterexample: $p=\infty$, $x=(1,0,0)$, $y=(\tfrac12,\tfrac12,\tfrac12)$.
(Visualize a cube centered at a vertex of an octahedron; the vertices of the cube (on the side toward the origin) overhang the facets of the octahedron, so a line from the origin to such a vertex of the cube will pass outside of the octahedron before it enters the cube.)

For the follow-up question: Take $p=\infty$ again, and take
\begin{align*}
  x &= (1,0,0,0,\dotsc) \\
  y &= (1,\underbrace{\varepsilon,\varepsilon,\dotsc,\varepsilon}_{\text{$n$ entries}},0,0,0,\dotsc)
\end{align*}
Then, if I've computed correctly, we have
$$ \left\|\frac{x}{\|x\|_1} - \frac{y}{\|y\|_1}\right\|_\infty = \frac{n}{1+n\varepsilon} \|x-y\|_\infty $$
so if we send $n\to\infty$ and $\varepsilon\to 0$ such that $n\varepsilon$ is constant, say, we'll get arbitrarily large stretching.
(At the level of informal visualization, what's happening here is that, for large $n$, the facets of the $\ell_1^n$ ball are very steep around the vertices, so projecting a point just slightly off from a vertex can take it a long way away.  This fits well intuitively with the (more familiar?) fact that the centres of the facets are very close to the origin.)
