Showing that $\text{exp}(\left \lvert z \right \rvert/(\left\lvert z \right\rvert - 1))\leq \left \lvert 1 + z \right\rvert$ Suppose $z \in \mathbb{C}$  with $\left\lvert{z}\right\rvert < 1$. I want to prove that that
$$\exp\left(\frac{|z|}{|z| - 1}\right)\leq |1 + z|.$$
I tried writing $z$ in cartesian form as $x + iy$ and in polar form as $re^{i\theta}$, but neither form seemed viable. I tried squaring the inequality, but that didn't seem useful either. Any ideas on how I can get started on this? 

Let $z = re^{i\theta}$, $0 < r < 1$, $\theta \in [0,2\pi]$.
The desired inequality becomes
$$
\exp\left(\frac{r}{r-1}\right) \leq |1+re^{i\theta}|.
$$
The RHS is:
$$
|1+re^{i\theta}| = \sqrt{(1+re^{i\theta})(1+re^{-i\theta})} = \sqrt{1+2r\cos\theta + r^2}.
$$
Now the desired inequality is equivalent to
$$
\frac{r}{r-1} \leq \frac12\ln(1+2r\cos\theta + r^2).
$$
For any $r$, the least the RHS can be is when $\theta = \pi$, so it is enough to show that
$$
\frac{r}{r-1} \leq \frac12\ln(1-2r + r^2) = \ln(1-r).
$$
The last inequality can be shown using calculus. 
