# Alternative proofs for $||z| - |w|| \leq |z + w|$ for $z, w \in \Bbb{C}$

One has to show that

$$||z| - |w|| \leq |z + w|; z,w \in \Bbb{C}$$

I proceeded as follows:

because of the symmetry on my next step, I can assume WLOG that $|z| \geq |w|$; Therefore $||z| - |w|| = |z| - |w|$:

$$|z| - |w|\leq |z + w| \iff\\ (|z| - |w|)^2 \leq |z+w|^2\iff \\ |z|^2 - 2|zw| + |w|^2 \leq (z+w)(\bar{z}+\bar{w})\iff\\ z\bar{z} + w\bar{w} - 2|zw| \leq z\bar{z} + w\bar{w} + z\bar{w} + \bar{z}w\iff \\ -2|zw| \leq 2Re(z\bar{w})$$

which is obviously true given that the LHS is $\leq 0$ and the RHS is $\geq 0$. I was wondering, however, if one could tackle this problem in a different way, making use of elementary tools only, to prove that inequality. I am particularly interested in knowing if there is any "obvious" way of applying the triangle inequality here.

• It is just a norm version of the triangle inequality. You can refer to the proof of the theorem: $||a|-|b||\leq|a+b|\leq|a|+|b|$, for all $a,b\in\mathbb{R}^2$. (it is usually presented in the more general $\Bbb R^n$ case.)
– Eric
Commented Jan 19, 2017 at 15:12
• Reverse triangle inequality is a general result for metric spaces and normed vector spaces. See a proof here and here
– edm
Commented Jan 19, 2017 at 15:15

Applying the triangle inequality to $|z|=|z+w-w|$ and $|w|=|w+z-z|$ yields $$|z|\leq |z+w| + |w| \Leftrightarrow |z| - |w| \leq |z+w|$$ and analogously $$|w| \leq |w+z| + |z| \Leftrightarrow |w| - |z| \leq |z+w|$$ which yields the wanted inequality.
We know :$$|w+z|\leq|w|+|z|$$ now put 1st $w \to w-z$ 2nd $z \to z-w$ $$1st\\|w+z|\leq|w|+|z| \to |(w-z)+z|\leq|w-z|+|z| \to |w|\leq |w-z|+|z| \to \\|w|-|z|\leq|w-z|\\2nd\\ |z|-|w|\leq|w-z|$$so now combine them $$|w-z| \geq |w|-|z|\\|w-z| \geq|z|-|w| \\\to |w-z| \geq||w|-|z||$$ at the end turn $z \to -z$ $$z \to -z \\ |w-(-z)| \geq||w|-|-z||\\|w+z| \geq||w|-|-1||z||\\|w+z| \geq||w|-|z||$$
• You ended up with $|w-z|$ on the LHS when the prob. statement is with $|w+z|$...