Show that $|z - w| \geq \big||z|-|w|\big|$ 
Show that $|z - w| \geq \big||z|-|w|\big|$ given $z = x + iy$ and $w = u + iv$. 

So far I have, 
$$|z - w| \geq \big||z|-|w|\big|$$
$$\sqrt{(x - u)^2 + (y - v)^2} \geq |\sqrt{x^2 + y^2} -\sqrt{u^2 + v^2}|$$
At this point I get a little 'undone'.
I tried a few different routes, but none of them have come up with the desired result.  

Other answers address your issue in a better way but I think it will be nice if a proof by your line of thought is given. 

$$\sqrt{(x - u)^2 + (y - v)^2} \geq |\sqrt{x^2 + y^2} -\sqrt{u^2 + v^2}|$$
Since both sides are $>0$,
$$(x - u)^2 + (y - v)^2 \geq x^2 + y^2 - u^2 + v^2  -2\sqrt{x^2 + y^2}\sqrt{u^2 + v^2}$$
$$(x^2 + u^2 -2xu) + (y^2 + v^2 - 2yv) \geq x^2 + y^2 + u^2 + v^2  -2\sqrt{x^2 + y^2}\sqrt{u^2 + v^2}$$
$$-xu - yv \geq -\sqrt{x^2 + y^2}\sqrt{u^2 + v^2}$$
$$\sqrt{x^2 + y^2}\sqrt{u^2 + v^2} -xu - yv \geq 0$$
$$\sqrt{x^2u^2 + x^2v^2 + y^2u^2 + y^2v^2} -xu - yv \geq 0$$
$$x^2u^2 + x^2v^2 + y^2u^2 + y^2v^2 \geq x^2u^2 + y^2v^2 + 2xyuv $$
$$x^2v^2 + y^2u^2  \geq  2xyuv \tag{*}$$
If you remember (*) is golden AM-GM inequality.
