Kanada, 2014

Note that for every positive integer $m$,
$$\begin{aligned} \frac{a_m}{(1+a_1)(1+a_2)\cdots(1+a_m)} &= \frac{1+a_m}{(1+a_1)(1+a_2)\cdots(1+a_m)} - \frac{1}{(1+a_1)(1+a_2)\cdots(1+a_m)} \\ &= \frac{1}{(1+a_1)\cdots(1+a_{m-1})} - \frac{1}{(1+a_1)\cdots(1+a_m)}. \end{aligned}$$
Therefore, if we let $b_j = (1+a_1)(1+a_2)\cdots(1+a_j)$, with $b_0 = 1$, then by telescoping sums,
$$\sum_{j=1}^{n} \frac{a_j}{(1+a_1)\cdots(1+a_j)} = \sum_{j=1}^{n} \left( \frac{1}{b_{j-1}} - \frac{1}{b_j} \right) = 1 - \frac{1}{b_n}.$$
Note that $b_n = (1+a_1)(1+a_2)\cdots(1+a_n) \ge (2\sqrt{a_1})(2\sqrt{a_2})\cdots(2\sqrt{a_n}) = 2^n$, with equality if and only if all $a_i$'s equal $1$. Therefore,
$$1 - \frac{1}{b_n} \ge 1 - \frac{1}{2^n} = \frac{2^n - 1}{2^n}.$$
To check that this minimum can be obtained, substitute all $a_i = 1$ to yield
$$\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots + \frac{1}{2^n} = \frac{2^{n-1} + 2^{n-2} + \dots + 1}{2^n} = \frac{2^n - 1}{2^n},$$
as desired.