my confusion arises when trying to derive $|a-b|\ge|a|-|b|$ My first attempt was to derive it using the triangle inequality i.e $|a-b| = |a+(-b)|\le|a|+|-b|=|a|+|b| since |a|=|-a|$ this is obviously wrong. So what I did next was try deriving it straight from first principles however I get two different answers when doing it from first principles
Frist Result: $$|a-b|^2=(a-b)^2=a^2-2ab+b^2\\ =|a|^2+|b|^2-2ab\ge |a|^2+|b|^2-2|a||b| =(|a|-|b|)^2\\ \implies|a-b|^2\ge(|a|-|b|)^2\implies|a-b|\ge|a|-|b| $$ wich is the correct answer. however we can do this In a different way
Second Result:
$$|a-b|^2=(a-b)^2 =a^2+2a(-b)+(-b)^2\\ =|a|^2+|-b|^2+2a(-b) \le|a|^2+|-b|^2+2|a||-b|\\ =(|a|+|-b|)^2=(|a|+|b|)^2 \textrm{ since } |-b|=|b| \implies|a-b|^2\le(|a|+|b|)^2\\ \implies|a-b|\le|a|+|b| $$ which is the result I got above in my first try.
Im not sure where im making a mistake in these calculations, however it seems like the two things contradict each other. Help would me much appreciated
Thank You
Your first attempt was in a wrong direction since when proving $$ |a-b|\geq |a|-|b|\tag{1} $$ you want a lower bound for the quantity $|a-b|$. But they way in your attempt gave an upper bound for $|a-b|$.
The fact that $$ |a-b|\leq |a|+|b|\tag{1'} $$ is not a contradiction to (1); it is nothing but another fact.
For example the fact that $|4-(-1)|\leq |4|+|-1|$ does not contradict $|4-(-1)|\geq |4|-|-1|=3$.
You may rewrite (1) as $$ |a|\leq |a-b|+|b|,\tag{2} $$ and if you apply the triangle inequality to $|(a-b)+b|$, you would get (2).