<s>[INST] If $a_1a_2\cdots a_n=1$, then the sum $\sum_k a_k\prod_{j\le k} (1+a_j)^{-1}$ is bounded below by $1-2^{-n}$ I am having trouble with an inequality.  Let $a_1,a_2,\ldots, a_n$ be positive real numbers whose product is $1$.  Show that the sum
 $$ \frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+\frac{a_3}{(1+a_1)(1+a_2)(1+a_3)}+\cdots+\frac{a_n}{(1+a_1)(1+a_2)\cdots(1+a_n)}$$
is greater than or equal to $$ \frac{2^n-1}{2^n}$$
If someone could help approaching this, that would be great. I don't even know where to start.
 [/INST] Note that  for that every positive integer $i$ we have 
\begin{eqnarray}
\frac{a_i}{(1+a_1)(1+a_2) \cdots (1+a_i)} & = & \frac{1 + a_i}{(1+a_1)(1+a_2) \cdots (1+a_i)} - \frac{1}{(1+a_1)(1+a_2) \cdots (1+a_i)} \nonumber \\
& = & \frac{1}{(1+a_1) \cdots (1+a_{i-1})} - \frac{1}{(1+a_1) \cdots (1+a_i)}. \nonumber
\end{eqnarray}
Let $b_i = (1+a_1)(1+a_2) \cdots (1+a_i)$, with $b_0= 0$. Then by telescopy
$$\sum\limits_{i=1}^n \left( \frac{1}{b_{i-1}} - \frac{1}{b_i} \right) = 1 - \frac{1}{b_n}.$$
Since $1+x\geq 2\sqrt{x}$ for all $x\ge 0$, we have
$$b_n = (1+a_1)(1+a_2) \cdots (1+a_n) \geq (2 \sqrt{a_1})(2 \sqrt{a_2}) \cdots (2 \sqrt{a_n}) = 2^n,$$
 with equality precisely if $a_i=1$ for all $i$. It follows that 
$$1 - \frac{1}{b_n} \geq 1 - \frac{1}{2^n} = \frac{2^n-1}{2^n}.$$
</s>