$|p\min(x,q)-y\min(x,w)| \leq x |p-y|?$ Let $x,y,w,p,q \geq 0$ such that $p,y \leq 1.$
Is it true that $$|p\min(x,q)-y\min(x,w)| \leq x |p-y|?$$
Tried to come with a proof but it didn't work, it seems the statement is false.
Any idea-example to disprove it?
 A: Since all variables are nonnegative (in case $y\ne 0$, $p=0$, exchange $p,y$ and $w,q$  in the following treatment; the case  $p=y=0$ is trivial) we can rewrite
$$
|p\min(x,q)-y\min(x,w)| \leq x |p-y|
$$
as
$$
|\frac{1-\frac{y/p}{\min(x,q) / \min(x,w)}}{1 -y/p}| \leq \frac{x}{\min(x,q)} 
$$
Denote $z = y/p$. As $0 \le y,p \le 1$ we have that $0\le z< \infty$.
Further, denote $a = \min(x,q) / \min(x,w)$. Then the question translates to
$$
|\frac{1-\frac{z}{a}}{1 -z}| \leq \frac{x}{\min(x,q)} 
$$

*

*First, consider $ a\ne 1$. This happens when $w > x \ge q$ or $q > x \ge w$. Then the LHS takes values from $0$ (for $z=a$) to $\to \infty$ (for $z=1$) so the LHS can, for all $z$, never be smaller than anything, which fails the assertion.


*Now, consider $ a =  1$. Then we have to show $1 \leq \frac{x}{\min(x,q)} $. This falls into two subcases.
2.1. $a = \min(x,q) / \min(x,w) =1$ is generated for $x >  w=q$. Then we have the condition $1 \leq \frac{x}{\min(x,q)} = \frac{x}{q} > 1$ which holds, with a generic "$<$" in the assertion.
2.2. $a = \min(x,q) / \min(x,w) =1$ is generated for $x \le q$, $x \le  w$. Then we have the condition $1 \leq \frac{x}{\min(x,q)} = \frac{x}{x}= 1$ which holds true, so in this  case  the assertion holds with equality.
A: For simply a counterexample, you could use $p=y=1$, $x=w=1$, $q=0$.
If the inequality was true, this would give $1\le0$.
