How to prove the following inequality $|\prod_{i=1}^{i=n}a_i-\prod_{i=1}^{i=n}b_i| < n\delta$?

Question

The constraints are

$0 \le a_1,a_2....a_n,b_1,b_2....b_n \le 1$.
$|a_i-b_i|< \delta$ for all $1 \le i \le n $

How do I go about proving the following $$|\prod_{i=1}^n a_i-\prod_{i=1}^n b_i| < n\delta$$

I tried reducing it to two terms where one term is like $(a-\delta)^n$ so that I can get the $n\delta$ term from binomial,but I am stuck. I would really appreciate just hints.

Induction over $n$ is an easy way to go. – Daniel Fischer Jun 29 '15 at 14:16 — Daniel Fischer, Jun 29 '15 at 14:16
@DanielFischer thanks, go it . – advocateofnone Jun 29 '15 at 14:24 — advocateofnone, Jun 29 '15 at 14:24

score 3 · Accepted Answer · answered Jun 29 '15 at 14:18

You can prove that:

$$\left|\prod\limits_{i=1}^{n}a_i-\prod\limits_{i=1}^{n}b_i\right|\le \sum\limits_{i=1}^{n} |a_i-b_i|.$$

The proof goes by induction. For $n=1$ it's obvious. Let's assume that it's true for some $n.$ We will prove this for $n+1.$ Let $A_n=\prod\limits_{i=1}^{n}a_i$, $B_n=\prod\limits_{i=1}^{n}b_i$. We can see that:

$$A_{n+1}-B_{n+1}=(a_{n+1}-b_{n+1})A_n+b_{n+1}(A_n-B_n).$$

The rest goes from the above identity, triangle inequality and from assumption about $a_i,b_i.$

How to prove the following inequality $|\prod_{i=1}^{i=n}a_i-\prod_{i=1}^{i=n}b_i| < n\delta$?

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