How to prove this inequality of bounded complex number.?

Question

$|a_1\cdots a_m - b_1\cdots b_m|\leq \sum\limits_{j=1}^m|a_i-b_i|$

Where $|a_i|\leq 1, |b_i|\leq 1$, and they are any complex number satisfying the condition.

Answer 1

First, let's prove the result for $m=2$. Aplying the triangle inequality:-

$$|a_1a_2 - b_1b_2|=|a_1(a_2-b_2)+b_2(a_1-b_1)|\le |a_1 - b_1|+|a_2 - b_2|.$$

Then $|a_1\cdots a_m - b_1\cdots b_m|$ can be written as $|(a_1\cdots a_{m-1})a_m - (b_1\cdots b_{m-1})b_m|$.

Applying the result for $m=2$ gives an upper bound of $$|(a_1\cdots a_{m-1}) - (b_1\cdots b_{m-1})|+|a_m - b_m|$$ and repeated application of this result gives the required inequality.

Answer 2

By induction on $m$.

For $m = 2$ we have:

$$\begin{aligned} \vert a_1 a_2 - b_1 b_2 \vert &= \vert a_1 a_2 -a_2 b_1 + a_2b_1 -b_1b_2 \vert\\ &\le \vert a_2 \vert \vert a_1 - b_1 \vert+ \vert b_1 \vert \vert a_2 - b_2 \vert\\ &\le \vert a_1 - b_1 \vert + \vert a_2 - b_2 \vert \end{aligned}$$

Induction step from $m$ to $m+1$ follows from the case $m=2$, substituting in previous proof $a_1 \dots a_m \to a_1$, $b_1 \dots b_m \to b_1$, $a_{m+1} \to a_2$ and $b_{m+1} \to b_2$.