# A third triangle inequality?

I was studying about complex numbers when I encountered this expression in my notebook, $$|a+b| \geq ||a|-|b||$$ It's different from the two triangle inequalities I already knew i.e. $$|a+b| \leq |a|+|b|$$ and $$|a-b| \geq ||a|-|b||$$ where $a$ and $b$ are any two complex numbers and $|.|$ represents the modulus function.

I couldn't find this inequality on the internet and even tried to prove it myself but don't know how to proceed.

So my question is

1. Is this expression right or I had just made some mistake copying it from the board?
2. If it's right, how to prove it?

Thanks for help:)

• It's no different from the first: just write $|a+b|=[a-(-b)|$ and note $|-b|=|b|$. – Bernard Mar 11 '18 at 19:37
• do you mean $$|a+b|\geq \left||a|-|b|\right|$$ or $$|a-b|\geq \left||a|-|b|\right|$$ – Dr. Sonnhard Graubner Mar 11 '18 at 19:37
• $|a+b| \geq ||a|-|b||$ and $|a-b| \geq ||a|-|b||$ are equivalent: just replace $b$ by $-b$. – Lord Shark the Unknown Mar 11 '18 at 19:38
• I am feeling like a fool now (T_T) – mayank mittal Mar 11 '18 at 19:40
• This is true for all norms, $| \|x\| - \|y\| | \le \| x-y\|$. – copper.hat Mar 11 '18 at 19:57

Take $-b$ instead of $b$ in your last equation (you can do that as your inequality holds for every complex number).
$$|\pm x\pm y|=|(\pm x)+(\pm y)|\ge||\pm x|-|\pm y||=||x|-|y||.$$