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.

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.
