Using reverse triangle inequality to prove inequality

Question

So I am trying to use the reverse triangle inequality on complex numbers, a,b, where $|a-b|>||a|-|b||$.

I need to use this to prove that if $|x| < |y|/2$, then $|x+y| >|y|/2$.

Intuitively this makes sense, and so I tried using the triangle inequality, so $|y-x| \geq ||y|-|x|| \geq ||y|-|y|/2| \geq ||y|/2| \geq |y|/2$ but this isn't right. Not sure how to approach or if I am thinking about this correctly. Any help would be appreciated!

Answer 1

Taking $a=x,b=-y$ we get $|x+y| \geq ||x|-|-y||= ||x|-|y||\geq |y|-|x|>|y|-\frac {|y|} 2=\frac {|y|}2$.

In the second inequality I have used the fact that $|t| \geq -t$ for any real number $t$.