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 ?
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 ?
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} $
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 $$ \prod_{i=1}^n(1-a_i) <\frac{1}{\prod_{i=1}^n(1+a_i)}<\frac{1}{1+\displaystyle \sum_{i=1}^n a_i} $$