If $\frac{1}{2}1-\left(a_1+\frac{a_2}{2}+\cdots+\frac{a_n}{2^{n-1}}\right)$ 
Let $n>1$ be a positive integer and $\frac{1}{2}<a_{j}<1$ for $j=1,2,\ldots,n$. Show that $$(1-a_{1})(1-a_{2})\cdots (1-a_{n})>1-\left(a_{1}+\frac{a_{2}}{2}+\cdots+\frac{a_{n}}{2^{n-1}}\right).$$

I have no ideas.
 A: Let $b_i=1-a_i$, so that $0<b_i<\frac{1}{2}$. The inequality becomes:
$$b_1b_2 \ldots b_n>1-((1-b_1)+\frac{1-b_2}{2}+\ldots+\frac{1-b_n}{2^{n-1}})$$
$$1-\frac{1}{2^{n-1}}+b_1b_2 \ldots b_n>b_1+\frac{b_2}{2}+\ldots+\frac{b_n}{2^{n-1}}$$
We first prove that $\frac{1}{2}+b_1b_2>b_1+\frac{b_2}{2}$. This is equivalent to $(\frac{1}{2}-b_1)(1-b_2)$, which is clearly true.
Now, we induct on $n \geq 2$ to prove the given inequality.
When $n=2$, we have already proven it above.
Suppose it holds for $n=k \geq 2$, and consider $k+1$. Using the induction hypothesis, (since $0<b_kb_{k+1}<\frac{1}{2}$.)
\begin{align}
1-\frac{1}{2^{k}}+b_1b_2 \ldots b_{k+1} & =\frac{1}{2^k}+(1-\frac{1}{2^{k-1}}+b_1b_2 \ldots b_{k-1}(b_kb_{k+1})) \\
& >\frac{1}{2^k}+b_1+\frac{b_2}{2}+\ldots +\frac{b_{k-1}}{2^{k-2}}+\frac{b_kb_{k+1}}{2^{k-1}}
\end{align}
Using the $n=2$ base case, 
$$\frac{1}{2^k}+\frac{b_kb_{k+1}}{2^{k-1}}=\frac{1}{2^{k-1}}(\frac{1}{2}+b_kb_{k+1})>\frac{1}{2^{k-1}}(b_k+\frac{b_{k+1}}{2})$$
Combining gives the desired inequality for $n=k+1$.
We are thus done by induction.
A: For another approach, we re-write as:
$$(1-a_1)(1-a_2) \dots (1-a_n) + \left(a_1 + \frac{a_2}{2} + \dots + \frac{a_n}{2^{n-1}}\right) > 1$$
Let $P = (1-a_1) (1-a_2)\dots (1-a_n)$ and $Q = \left(a_1 + \frac{a_2}{2} + \dots + \frac{a_n}{2^{n-1}}\right) $.  It may be noted that for any $a_k$ we can define $p = \dfrac{P}{1-a_k}$ which does not depend on $a_k$ and $0 < p < \dfrac{1}{2^{n-1}}$.  Similarly $q = Q - \frac{a_k}{2^{k-1}}$ does not depend on $a_k$.
Now $LHS = f(a_k) = P + Q = p(1-a_k) + q + \dfrac{a_k}{2^{k-1}} = \left(\dfrac{1}{2^{k-1}}-p\right)a_k + p+q$
is a linear function of $a_k$, with positive slope $\frac{1}{2^{k-1}}-p$.  
Hence $f(a_k)$ achieves its lower bound at the lower limit of its domain, i.e. when $a_k \rightarrow \frac{1}{2}$.   As the reasoning was for a general $a_k$, the minimum of the LHS must be when all $a_k \rightarrow \frac{1}{2}$.  Hence $LHS > \dfrac{1}{2^n} + (1 - \dfrac{1}{2^n}) = 1$.  
EDITED to explain the general case and clarify the notations used.
A: An obvious approach would be induction.
However, it is not true for $n=1$,
so let's use $\ge$ instead of $>$.
For $n=1$ it is an identity.
For $n=2$, we want
$(1-a_1)(1-a_2) \ge 1-a_1-a_2/2$
or
$1-a_1-a_2+a_1 a_2 \ge 1-a_1-a_2/2$
or 
$a_1 a_2 \ge a_2/2$
or
$a_1 \ge 1/2$
which is true by hypothesis.
So, it might be true.
(just noticed that there is a new solution.
Be interesting to see how these compare.)
Suppose that
$\prod_{i=1}^n (1-a_i) \ge 
1-\sum_{i=1}^n a_i/2^{i-1}$.
Let's multiply both sides by $1-a_{n+1}$.
We get
$\begin{align}
\prod_{i=1}^{n+1} (1-a_i) 
&\ge 
(1-\sum_{i=1}^n a_i/2^{i-1})(1-a_{n+1})\\
&=1-\sum_{i=1}^n a_i/2^{i-1}-
a_{n+1}(1-\sum_{i=1}^n a_i/2^{i-1})
\end{align}
$.
To get the induction to work, we need
$a_{n+1}(1-\sum_{i=1}^n a_i/2^{i-1})
\le a_{n+1}/2^n$
or
$1-\sum_{i=1}^n a_i/2^{i-1} \le 1/2^n$.
But, since $a_i > 1/2$,
$\sum_{i=1}^n a_i/2^{i-1}
> \sum_{i=1}^n 1/2^{i}
= 1-1/2^{n}
$
which is exactly what we need.
