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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:)

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  • $\begingroup$ 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
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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.

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No, it's not different.

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

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