Is there a tighter bound on $\prod_{i=1}^I x_i$?

Question

If $x_i\in[0,1]$ for all $i=1,2,\cdots,I$, then we have a simple bound on $\prod_{i=1}^Ix_i$, i.e., $$\prod_{i=1}^I x_i \le \sum_{i=1}^I x_i.$$

I wonder if there is a tighter bound than the one above?

Answer 1

The tightest you can get bounding by one element is $\bar{x} = \min\{x_i : i \in \{1, \ldots, I\}\}$, since a multiplication by a number less than 1 will lower the product. Hence, $$\prod_{i=1}^I x_i \leq \bar{x} $$ But of course, you can always bound multiplying by one element less $$\prod_{i=1}^I x_i \leq \prod_{i=2}^I x_i \leq \cdots$$

Answer 2

The Arithmetic Mean Geometric Mean (AM-GM) inequality says that $$\dfrac{1}{I}\sum_{i = 1}^{I}x_i \ge \left(\prod_{i = 1}^{I}x_i\right)^{1/I},$$ and thus, we get the bound $$\prod_{i = 1}^{I}x_i \le \left(\dfrac{1}{I}\sum_{i = 1}^{I}x_i\right)^I.$$

We can check that this bound less than or equal to $\displaystyle\sum_{i = 1}^{I}x_i$ for all $x_i \in [0,1]$.