Upper bound on the distance of two products Let $(a_1,\ldots,a_n)\in[0,1]^n$ and $(b_1,\ldots,b_n)\in[0,1]^n$.
Let $\epsilon\in[0,1]$ and assume $\forall i:|a_i-b_i|\leq\epsilon$ and $\forall i,j:|a_i-a_j|\leq\epsilon$.
Question:
Is $\left|\prod_{i\in[n]}a_i-\prod_{i\in[n]}b_i\right|\leq n\epsilon$?
(A proof would be appreciated.)
 A: Here's a proof . .  .

Fix a positive integer $n$, and let $\epsilon \in [0,1]$.

Suppose we have a pair $(a,b)$, with $a,b\in [0,1]^n$ such that
\begin{align*}
&{\small{\bullet}}\;\;|a_i - b_i|\le \epsilon,\;\text{for all}\;i\\[7pt]
&{\small{\bullet}}\;\;|a_i - a_j|\le \epsilon,\;\text{for all}\;i,j\\[3pt]
&{\small{\bullet}}\;\;\left|\prod_{i=1}^n a_i-\prod_{i=1}^n b_i\right| > n\epsilon\\[3pt]
\end{align*}
In other words, assume the pair $(a,b)$ is a counterexample to the claim.

Without loss of generality, we can assume
$$a_1 \le \cdots \le a_n$$
Since, by assumption,
$$\left|\prod_{i=1}^n a_i-\prod_{i=1}^n b_i\right| > n\epsilon$$
we can't have 
$$\prod_{i=1}^n b_i=\prod_{i=1}^n a_i$$
Consider two cases . . .

Case $(1)$:$\;\;{\displaystyle{\prod_{i=1}^n b_i > \prod_{i=1}^n a_i}}$.

Given any such counterexample, we can replace it with one satisfying
$$b_i = \min(1,a_i+\epsilon),\;\text{for all}\;i$$
Thus, assume the above condition is satisfied.

It follows that $b_1 \le \cdots \le b_n$. 

Suppose $b_1 < b_n$. 

Then, letting $w=b_n-b_1$, we can replace 
\begin{align*}
&{\small{\bullet}}\;\;a_1\;\text{by}\;a_1+w\\[4pt]
&{\small{\bullet}}\;\;b_1\;\text{by}\;b_1+w=b_n\\[4pt]
\end{align*}
and we would still have a case $(1)$ counterexample.

Iterating these replacements, we can get a case $(1)$ counterexample for which
\begin{align*}
&{\small{\bullet}}\;\;b_1 = \cdots = b_n = x\\[4pt]
&{\small{\bullet}}\;\;a_1 = \cdots = a_n = y\\[4pt]
\end{align*}
where $0 \le x-y \le \epsilon$.

Then, noting that $0 \le x,y \le 1$, we get

\begin{align*}
\left|\prod_{i=1}^n a_i-\prod_{i=1}^n b_i\right| 
&=\prod_{i=1}^n b_i-\prod_{i=1}^n a_i\\[4pt]
&=x^n-y^n\\[4pt]
&=(x^{n-1}+x^{n-2}y + \cdots + xy^{n-2}+y^{n-1})(x-y)\\[4pt]
&\le(x^{n-1}+x^{n-2}y + \cdots + xy^{n-2}+y^{n-1})\epsilon\\[4pt]
&\le n\epsilon
\end{align*}
contradiction.

Hence, for case $(1)$, no counterexample is possible.

Case $(2)$:$\;\;{\displaystyle{\prod_{i=1}^n a_i > \prod_{i=1}^n b_i}}$.

Given any such counterexample, we can replace it with one satisfying
$$b_i = \max(0,a_i-\epsilon),\;\text{for all}\;i$$
Thus, assume the above condition is satisfied.

It follows that $b_1 \le \cdots \le b_n$. 

Suppose $b_1 < b_n$. 

Then, letting $w=b_n-b_1$, we can replace 
\begin{align*}
&{\small{\bullet}}\;\;a_1\;\text{by}\;a_1+w\\[4pt]
&{\small{\bullet}}\;\;b_1\;\text{by}\;b_1+w=b_n\\[4pt]
\end{align*}
and we would still have a case $(2)$ counterexample.

Iterating these replacements, we can get a case $(2)$ counterexample for which
\begin{align*}
&{\small{\bullet}}\;\;a_1 = \cdots = a_n = x\\[4pt]
&{\small{\bullet}}\;\;b_1 = \cdots = b_n = y\\[4pt]
\end{align*}
where $0 \le x-y \le \epsilon$.

Then, noting that $0 \le x,y \le 1$, we get

\begin{align*}
\left|\prod_{i=1}^n a_i-\prod_{i=1}^n b_i\right| 
&=\prod_{i=1}^n a_i-\prod_{i=1}^n b_i\\[4pt]
&=x^n-y^n\\[4pt]
&=(x^{n-1}+x^{n-2}y + \cdots + xy^{n-2}+y^{n-1})(x-y)\\[4pt]
&\le (x^{n-1}+x^{n-2}y + \cdots + xy^{n-2}+y^{n-1})\epsilon\\[4pt]
&\le n\epsilon
\end{align*}
contradiction.

Hence, for case $(2)$, no counterexample is possible.

Thus, in both cases, no counterexample is possible, which completes the proof of the claim.
