An Inequality Related to Absolute Values

Question

There exist two very known inequalities related to absolute values of complex numbers $x$ and $y$, namely, $|x|-|y|\leq |x+y|\leq |x|+|y|$.

However, I would like to find a kind of reverse inequality with possible constants and with some (which can be useful) hypothesis. In fact, my question is:

Question. Does there exist positive constants $k$ and $r$, such that for any complex numbers $x$ and $y$, with $|x|>|y|>2$, the estimate $$ |x|-|y|\geq k(|x-y|)^r $$ holds?

Any suggestion?

Note that if $x$ and $y$ are positive real numbers, then $x=y+\delta$, for some positive $\delta$. Then, $$ |x|-|y|=\delta\geq \delta=k(|x-y|)^r, $$ for $k=r=1$.

Answer 1

No such inequality exists. Take $x=1+\epsilon$ and $y=-1$. The inequality fails for $\epsilon$ sufficiently small (whatever $k$ and $r$ may be).

Answer 2

Let $y = -x + \delta$ where $\operatorname{arg}x = \operatorname{arg}\delta$, then $|x| - |y| = |\delta|$

$\forall k,r > 0\ \ k |x - y|^r \ge k$ if $|x - y| \ge 1$

$\therefore \exists \delta \ \text{s.t} \ |\delta| < k$

$\therefore \forall k,r > 0\ \exists x,y\ \text{s.t.} |x| - |y| < k |x - y|^r$ where |x| > |y| > 2