Let $x,y,z$ be distinct complex numbers such that $$y=tx+(1-t)z$$ ; where $t \in (0,1)$
Prove that $$\frac{|z|-|y|}{|z-y|} \geq \frac{|z|-|x|}{|z-x|} \geq \frac{|y|-|x|}{|y-x|} $$
I tried this question by finding $t$ in terms of $x,y,z$. But I'm getting an inequality which doesn't lead to the answer.
$$|t|= \frac{|y-z|}{|x-z|} $$
So we have $$0 \leq|y-z|\leq|x-z|$$
Now it is clear that $$0 \leq|y|-|z|\leq|x-z|$$
After this ? I;m stuck. Any ideas ?
Note that $|z-y|=|-tx+tz|=t|z-x|$. Meanwhile, by the triangle inequality, $$|z|-|y|\geq|z|-t|x|-(1-t)|z|=t(|z|-|x|)$$ and so we now have $$\frac{|z|-|y|}{|z-y|}\geq\frac{t(|z|-|x|)}{t|z-x|}=\frac{|z|-|x|}{|z-x|}.$$ The remaining inequality should be similar.