Proof of inequality with absolute values used in Ishii/Sato 04 and many PDE existence proofs In Ishii/Sato 2004 Page 1095, Gout 2005 Page 169, Guillot 2016 Page 17 they use that 
$
\left| \frac{p}{|p|} - \frac{q}{|q|}\right|^{2} \le 4\rho(p,q)^{2}
$
where $p,q\in\mathbb{R}^{n}\backslash\{ 0\}$ and
$
\rho(p,q) = \min\left(\frac{|p-q|}{\min(|p|,|q|)},1\right).
$
The inequality $\left| \frac{p}{|p|} - \frac{q}{|q|}\right|^{2} \le \frac{|p-q|}{\min(|p|,|q|)}$ seems important, and some writers confusingly define $\rho(p,q)$ as above, but in the same paper will define it $\rho(p,q) = \frac{|p-q|}{\min(|p|,|q|)}$ (making the first result obvious).
However, it is crucial in Sato based proofs that $\rho$ is defined as at the top, and therefore I cannot see how it still holds.
Ishii/Sato quote an example in $\S 5$ of their paper with the inequality
$
\left| \frac{p}{|p|} - \frac{q}{|q|}\right| \le \frac{|p-q|}{\max(|p|,|q|)},
$
if this holds, then the problem is over - however I'm not certain that it is and cannot find a proof of it.

Let $p= \alpha x$, $q = \beta y$, with $\alpha = |p|>0$, $\beta = |q|>0$, $|x|= |y|=1$, so that the inequality becomes
$$
|x-y|^2 \leq 4 \left[\min\left(\frac{|\alpha x - \beta y|}{\min\{\alpha, \beta\}}\,, 1\right)\right]^2\,.
$$
Since $|x-y|\leq 2$, clearly it is enough to prove that
$$
|x-y|^2 \leq 4 \left[\frac{|\alpha x - \beta y|}{\min\{\alpha, \beta\}}\right]^2\,.
$$
It is not restrictive to assume that $0< \alpha\leq\beta$, so that $\lambda := \beta / \alpha \geq 1$ and the inequality becomes
$$
|x-y|^2 \leq 4 |x-\lambda y|^2
\qquad
(|x|=|y|=1,\ \lambda\geq 1).
$$
Let us consider the function
$$
\varphi(\lambda) := |x-y|^2 - 4 |x-\lambda y|^2.
$$
We have that $\varphi(1) = - 3 |x-y|^2 \leq 0$, and
$$
\varphi'(\lambda) = 8 x\cdot y - 8 \lambda |y|^2 \leq 8(1-\lambda) < 0,
\quad \forall \lambda > 1,
$$
so that $\varphi$ is strictly decreasing in $[1,+\infty)$.
Since $\varphi(1) \leq 0$, we conclude that $\varphi(\lambda) < 0$ for every $\lambda > 1$.
