Complex-number inequality $| z_1 z_2 \ldots z_m - 1 | \leq e^{|z_1 - 1| + \ldots + |z_m - 1|} - 1$ Let $z_1, z_2 \ldots z_m$ be complex numbers, $m \in \mathbb{N}$. Can anybody tell me how to prove the following inequality?
$| z_1 z_2 \ldots z_m - 1 | \leq e^{|z_1 - 1| + \ldots + |z_m - 1|} - 1$
In case you're wondering, this is asserted without proof in a paper by Von Neumann, about infinite tensor products of Hilbert spaces.
 A: The inequality in question bounds how far you can get from $1$ by multiplying several complex numbers that may individually not be far from $1$. So it makes sense to try to derive a bound for the product of just two complex numbers, and then proceed by induction.
Lemma: Suppose $\lvert z_1 - 1\rvert = \alpha_1$ and $\lvert z_2 - 1\rvert = \alpha_2$. Then $\lvert z_1 z_2 - 1\rvert \le (1 + \alpha_1)(1 + \alpha_2) - 1$.
Proof: We know $\alpha_1\alpha_2 = \lvert z_1z_2 - z_1 - z_2 + 1\rvert$. By triangle inequality on the three points $z_1 z_2$, $z_1 + z_2 - 1$, and $1$, we have
$$\begin{align}
\lvert z_1 z_2 - 1\rvert &\le \lvert z_1 z_2 - z_1 - z_2 + 1\rvert + \lvert z_1 + z_2 - 2\rvert \\
&\le \alpha_1\alpha_2 + \alpha_1 + \alpha_2 \\
&= (1 + \alpha_1)(1 + \alpha_2) - 1.
\end{align}$$
Now, for several numbers,
$$\begin{align}
\lvert z_1 z_2 \cdots z_m - 1\rvert &\le (1 + \alpha_1)(1 + \alpha_{2,\ldots,m}) - 1 \\
&\le (1 + \alpha_1)(1 + \alpha_2)(1 + \alpha_{3,\ldots,m}) - 1 \\
&\vdots \\
&\le (1 + \alpha_1)(1 + \alpha_2)\cdots(1 + \alpha_m) - 1,
\end{align}$$ where $\alpha_{2,\ldots,m}$, for example, is my hopefully transparent abuse of notation to denote $\lvert z_2\cdots z_m - 1\rvert.$ Finally, since $1+x \le e^x$ for real $x$, the desired inequality follows.
A: EDIT: This answer is wrong.
It all boils down to the inequality $|xy-1| \leq |x-1|+|y-1|$, which I expect to be true.
Given this inequality, prove by induction that
$|z_1\cdots z_m - 1| \leq |z_1 - 1| + \cdots + |z_m - 1|$.
Now use $e^x \geq 1 + x$.
