<s>[INST] 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.
 [/INST] 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.
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