Complex-number inequality |z1z2⋯zm−1|≤ez1−1|+⋯+|zm−1∣−1 Can anybody tell me how to prove the following inequality?

Question

Complex-number inequality  |z1z2⋯zm−1|≤ez1−1|+⋯+|zm−1∣−1  Can anybody tell me how to prove the following inequality?

nick1337 · Accepted Answer

Step 1 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 |z1−1|=α1 and |z2−1|=α2. Then |z1z2−1|≤(1+α1)(1+α2)−1 Proof: We know α1α2=|z1z2−z1−z2+1|. By triangle inequality on the three points z1z2, z1+z2−1, and 1, we have |z1z2−1|≤|z1z2−z1−z2+1|+|z1+z2−2| ≤α1α2+α1+α2 =(1+α1)(1+α2)−1 Now, for several numbers, |z1z2⋯zm−1|≤(1+α1)(1+α2,⋯,m)−1 ≤(1+α1)(1+α2)(1+α3,⋯,m)−1 ⋮ ≤(1+α1)(1+α2)⋯(1+αm)−1 where a2,⋯,m, for example, is my hopefully transparent abuse of notation to denote |z2⋯zm−1|. Finally, since 1+x≤ex for real x, the desired inequality follows.

Vasquez · Accepted Answer

EDIT: This answer is wrong. It all boils down to the inequality |xy−1|≤|x−1|+|y−1|, which I expect to be true. Given this inequality, prove by induction that |z1⋯zm−1|≤|z1−1|+⋯+|zm−1| Now use ex≥1+x

Complex-number inequality |z_{1}z_{2}\cdots z_{m}-1|\leq e^{z_{1}-1|+\cdots+|z_{m}-1|}-1 Can anybody tell me how to prove

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