$|z-1|\leq \bigg| |z|-1\bigg|+|z| |Arg\:z|$

For a non-zero $z$ and $-\pi<Arg\:z\leq \pi$, show that $$|z-1|\leq \bigg| |z|-1\bigg|+|z| |Arg\:z|.$$

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$$\begin{align}|z-1|&=\bigg|z-|z|+|z|-1\bigg|\\
&\leq \bigg|z-|z|\bigg|+\bigg||z|-1\bigg|\\
&=|z||\cos\theta+i\sin\theta-1|+\bigg||z|-1\bigg|\\
&=|z|\sqrt{(\cos\theta-1)^2+\sin^2\theta}+\bigg||z|-1\bigg|\\
&=|z|\left|2\sin\frac{\theta}{2}\right|+\bigg||z|-1\bigg|\\
&\leq |z||Arg\:z|+\bigg||z|-1\bigg|\end{align}$$ since $(\cos\theta-1)^2+\sin^2\theta=(1-2\sin^2\dfrac{\theta}{2}-1)^2+4\sin^2\dfrac{\theta}{2}\cos^2\dfrac{\theta}{2}=4\sin^2\dfrac{\theta}{2}$ and $\bigg|\sin\dfrac{\theta}{2}\bigg|<\bigg|\dfrac{\theta}{2}\bigg|$.

Geometric interpretation

Take $|z|>1$. Consider the triangle with vertices $1,z,|z|$. By the triangle inequality, we have $|z-1|\leq \bigg|z-|z|\bigg|+\bigg||z|-1\bigg|$. Recall the arc length formula $s=r\theta$. Here, $|z| |Arg\:z|$ is actually the arc length joining $z$ and $|z|$, which is larger than the chord joining the two points. Therefore, the inequality holds.