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:)

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

This means that the last inequality is equivalent to the one you're asking about.


No, it's not different.

$$|\pm x\pm y|=|(\pm x)+(\pm y)|\ge||\pm x|-|\pm y||=||x|-|y||.$$


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