Prove that $1-\sum_{i=1}^n a_i < \prod_{i=1}^n(1-a_i) $ 
Let $a_1, a_2, a_3,\ldots a_n$ be positive real numbers where $n > 1$. Prove that $$1-\displaystyle \sum_{i=1}^n a_i < \prod_{i=1}^n(1-a_i) $$

Can this be proved using the binomial theorem ?
 A: For some of the steps
in the following proof by induction
to be true,
there needs to be
some restrictions
on the $a_i$.
Try to find what is needed
in order to make the proof valid.
If
$\prod_{i=1}^n(1-a_i)
\gt 1-\sum_{i=1}^n a_i  
$
then
$\begin{array}\\
\prod_{i=1}^{n+1}(1-a_i)
&=(1-a_{n+1})\prod_{i=1}^{n}(1-a_i)\\
&\gt (1-a_{n+1})(1-\sum_{i=1}^n a_i)\\
&=1-a_{n+1}-\sum_{i=1}^n a_i+a_{n+1}\sum_{i=1}^n a_i\\
&=1-\sum_{i=1}^{n+1} a_i+a_{n+1}\sum_{i=1}^n a_i\\
&>1-\sum_{i=1}^{n+1} a_i\\
\end{array}
$
A: You can prove $1-\displaystyle \sum_{i=1}^n a_i< \prod_{i=1}^n(1-a_i)$ using simple induction.
I can prove the inequality $\displaystyle \prod_{i=1}^n(1-a_i)<\frac{1}{1+\sum_{i=1}^na_i}$ assuming that $\forall i, a_i<1$.
Indeed, you can prove first in a similar fashion that $1+\displaystyle \sum_{i=1}^n a_i< \prod_{i=1}^n(1+a_i)$.
Notice then that $\displaystyle \prod_{i=1}^n(1-a_i)(1+a_i)=\prod_{i=1}^n(1-a_i^2) < 1$, hence $$ \prod_{i=1}^n(1-a_i) <\frac{1}{\prod_{i=1}^n(1+a_i)}<\frac{1}{1+\displaystyle \sum_{i=1}^n a_i}  $$
