Kanada 2014

Note that for every positive integer $m$, $$\begin{array}{rl}\frac{a_m}{(1+a_1)(1+a_2)\cdots (1+a_m)}& =\frac{1+a_m}{(1+a_1)(1+a_2)\cdots (1+a_m)}-\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_m)}\\ & =\frac{1}{(1+a_1)\cdots (1+a_{m-1})}-\frac{1}{(1+a_1)\cdots (1+a_m)}.\end{array}$$ 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 $$\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots +\frac{1}{2^n}=\frac{2^{n-1}+2^{n-2}+\cdots +1}{2^n}=\frac{2^n-1}{2^n},$$ as desired.