Let $z=a+ib,w=c+id\in\mathbb{C}. $Prove $|z|-|w|\leq|z-w|$
Solution: If $|z|-|w|\leq 0$ then we are done since $|z-w|\geq 0$. Suppose $|z|-|w|>0$. The stages are going backward: $$ \\ |z|-|w|\leq|z-w|\Leftarrow \\ (|z|-|w|)^2\leq |z-w|^2\Leftarrow \\ (\sqrt{a^2+b^2}-\sqrt{c^2+d^2})^2\leq(a-c)^2+(b-d)^2\Leftarrow \\ ac+bd\leq c^2+d^2+\sqrt{(a^2+b^2)(c^2+d^2)} $$
How can I finish it?
Also, is there a way to solve it without using $a,b,c$ and $d$?
On
Note that $|z|-|w|\le||z|-|w||$ from properties of the absolute value. Now
$$\begin{align} ||z|-|w||\le|z-w| &\iff(|z|-|w|)^2\le|z-w|^2\\ &\iff|z|^2-2|zw|+|w|^2\le(z-w)\overline{(z-w)}=|z|^2-(z\overline{w}+\overline{z}w)+|w|^2\\ &\iff{z\overline{w}+\overline{z}w\over2}\le|zw|\\ &\iff\Re(z\overline{w})\le|zw| \end{align}$$
But $|zw|=|z\overline{w}|=\sqrt{(\Re(z\overline{w})^2+(\Im(z\overline{w})^2}\ge\Re(z\overline{w})$, so the inequalities above all hold. Thus
$$|z|-|w|\le||z|-|w||\le|z-w|$$
gives the desired result.
From triangle inequality, $$ \\ |z|=|z-w+w|\leq |z-w|+|w|\Rightarrow \\ |z|-|w|\leq |z-w| $$