\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.
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