Why is $|x-y|^2 \geq \frac 12 |y|^2$ when $|x|$ is sufficiently small?

Question

Why is $|x-y|^2 \geq \frac 12 |y|^2$ when $|x|$ is sufficiently small. Here $x,y \in \Bbb R^n$.

Logically it seems obvious because as $|x|$ is small $|x-y|$ is eventually very close to $|y|$ but I can't prove it using Triangle inequality. Maybe I am missing something. The furthest I went;

$$||x|-|y||\leq |x-y| \Rightarrow -|x-y| \leq |x|-|y|\leq |x-y|$$ If we take $|x|\leq \frac{|y|}{2}$ then $$\Rightarrow \frac{|y|}{2} -|y|\leq |x|-|y|\leq |x-y|$$ $$\Rightarrow -\frac{|y|}{2}\leq |x-y|$$ but still I can't conclude the result as $-\frac{|y|}{2}<0$.

Answer 1

If $|x|<\varepsilon|y|,$ then: $$|y-x|^2\geq ||y|-|x||^2=(|y|-|x|)^2>|y|^2(1-\varepsilon)^2$$ for a small $\varepsilon.$ So you can even quantify what exactly is sufficiently small in terms of this $\varepsilon.$