# Absolute value of product is less than product of absolute values: $|(1+a_1)(1+a_2)\dots (1+a_n)-1|\leq (1+|a_1|)(1+|a_2|)\dots (1+|a_n|)-1$

For a sequence $a_n\in\mathbb{C}$ I want to show that $$|(1+a_1)(1+a_2)\dots (1+a_n)-1|\leq (1+|a_1|)(1+|a_2|)\dots (1+|a_n|)-1$$

I think I should show this by induction on $n$. For the base case I'm sort of stuck, because $|(1+a_1)(1+a_2)-1|=|1+a_1+a_2+a_1a_2-1|\leq 1+|a_1|+|a_2|+|a_1a_2-1|$ and I don't see how to factor this into the result I want. The inductive step also isn't clear after I assume the statement to be true for some $n>2$. There might be some property of $\mathbb{C}$ that I'm missing for this.

• The $n=2$ case looks more like an inductive case. I would start with the case $n=1$ as the base case. This, however, doesn't address your question... – Michael Burr Oct 5 '15 at 3:01

$$|(1+a_1)(1+a_2)-1|=|a_1+a_2+a_1a_2|\leq|a_1|+|a_2|+|a_1a_2|=(1+|a_1|)(1+|a_2|)-1$$ by the triangle inequality. This generalizes for all products without induction.