Show $|x-y'|\leq |x-y|$ where $0<|x|\leq |y|<1$ and $y'=(1-|x-y|)y/|y|$

Question

Let $x,y \in \mathbb C$ with $0<|x|\leq |y|<1$. Put $\delta=|x-y|$, then we see that $\delta \in [0,2)$ and also that $1-\delta \in (-1,1]$. Define

$$ y'=\frac{1-\delta}{|y|}y.$$

Assume $1-|y|<\delta \leq 1-|x|$, show $|x-y'|\leq |x-y|$.

This is a step which is omitted in a proof, and I (unfortunately) dont get my algebra straight.

Answer 1

A "loose" proof:

You have $$ 1 - \lvert y \rvert < \delta \le 1 - \lvert x \rvert \implies \lvert x \rvert \le 1 - \delta < \lvert y \rvert \implies \lvert x \rvert \le \lvert y' \rvert < \lvert y \rvert, $$

also $y$ and $y'$ lie on the same ray from origin. Consider the diagram (point $X$ represents $x \in \mathbb{C}$ etc.):

Now,

1. $m(\angle OY'X) \le \frac{\pi}{2}$ ($X$ is "inside" the middle circle, ...)
2. $m(\angle XY'Y) \ge \frac{\pi}{2}$, so normal from $X$ on $OY$ drops inbetween $O$ and $Y'$
3. Express $XY$ in terms of length of the normal and cosine of the angle sustained by $XY$ w.r.t. the normal. Same for $XY'$. Cosine decreases as angle increases.