Min/max inequality

Question

Since, we have $$|x|=|x-y+y|\leq |x-y|+|y|,$$ then the following holds $$|x|\leq 2\max( |x-y|,|y|).$$ My question is: do we have also the following inequality $$\frac{1}{2}\min( |x-y|,|y|)\leq |x|?$$ if so, would you please sketsh the proof?

Answer 1

Let $y=-kx$ for $k>2$ then we have

$$ \frac{1}{2} \min(|x-y|,|y|)= \frac{kx}{2}>x$$

Answer 2

Let $x=1$ and $y=1000$ then $$1/2\min(|x-y|,|y|)=999/2>1=|x|$$