# Prove the following based on the triangle inequality

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:

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

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

$$0\le|A+B|\le |A|+|B|$$ and
$$0<||C|-|D||\le |C+D|$$
$$|A+B|\cdot||C|-|D||\le (|A|+|B|)|C+D|$$ and since $|C|\ne |D|$ then $C\ne\pm D$ hence $C+D\ne0$ and the result follows by dividing by $||C|-|D||\cdot|C+D|$.