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So I've just proven $|z_1+z_2|\le |z_1|+|z_2|$

and then I proved

if $a=z_1+z_2$ and $b=z_2$ then $||a|-|b||\le|a-b|$

And now I have to prove the following:

enter image description here

You can see that the top half is the triangle inequality... but then the bottom is the opposite of the second thing I had to prove. I had a friend that can prove this, only if it's based on the fact that $C\gt D$ but she didn't prove it in the other case where $D\gt C$ so I don't understand what to do.

Thanks

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  • $\begingroup$ Try replacing $b$ with $-b$ and you get the denominator. $\endgroup$ – Gregory Grant Feb 23 '15 at 13:03
  • $\begingroup$ Hint: if one divides with a smaller number gets bigger quotient. $\endgroup$ – Janko Bracic Feb 23 '15 at 13:05
  • $\begingroup$ @GregoryGrant Thanks, I didn't spot that but then the inequality sign is the wrong way round. $\endgroup$ – Douglas Feb 23 '15 at 13:06
  • $\begingroup$ @JankoBracic and sorry, I don't understand what you're trying to say? $\endgroup$ – Douglas Feb 23 '15 at 13:06
  • $\begingroup$ You need to prove the second inequality in general. You have only proved it for particular values of $a$ and $b.$ $\endgroup$ – Krish Feb 23 '15 at 13:08
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We have

$$0\le|A+B|\le |A|+|B|$$ and

$$0<||C|-|D||\le |C+D|$$

so

$$|A+B|\cdot||C|-|D||\le (|A|+|B|)|C+D|$$ and since $|C|\ne |D|$ then $C\ne\pm D$ hence $C+D\ne0$ and the result follows by dividing by $||C|-|D||\cdot|C+D|$.

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  • $\begingroup$ Thanks a lot. This is well explained and well justified!! $\endgroup$ – Douglas Feb 23 '15 at 13:15
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Feb 23 '15 at 13:17

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