$|z-w| \leq |w| $ implies $|z+w| \geq |w| $?

Question

If $z$ and $w$ are complex numbers such that $$|z-w| \leq |w| $$ is it true that $|z+w| \geq |w| $ ?

I believe this is true but I don't know how to prove it. Any help would be very appreciated. Thank you!!

Answer 1

Recall the parallelogram identity: $$|z - w|^2 + |z + w|^2 = 2|z|^2 + 2|w|^2.$$ Hence, $$|z - w|^2 + |z + w|^2 \ge 2|w|^2.$$ Therefore, at least one of $|z - w|^2$ or $|z + w|^2$ must be greater than or equal to $|w|^2$, which is what you need.

Answer 2

$|z+w|\geq 2|w|-|z-w|\geq |w|$.

Answer 3

$$|z + w| \ge |2w| - |w - z| \ge 2|w| - |w| = |w|.$$

Answer 4

Suppose $|z-w|\leq|w|$ so that $-|z-w|\ge-|w|$. Then, \begin{multline*} \left|z+w\right| =\left|2w+z-w\right| =\left|2w-\left(w-z\right)\right| \geq2\left|w\right|-\left|w-z\right| \\=2\left|w\right|-\left|z-w\right| \geq2\left|w\right|-\left|w\right| =\left|w\right| \end{multline*}