Let $n\geq 1$ be an integer and let $a_1,\ldots,a_n$ be positive real numbers, all between $0$ and $1$.
Is it possible to prove or disprove: $$ {(\prod_{i=1}^{n}(1-a_i))}{(1+\sum_{i=1}^{n}a_i)}<1 $$ To prove this, I was able to prove this to be true if all $a_1,\ldots,a_n$ are the same which is$(1-a)^n(1+an)<1 $ using Bernoulli's Inequality:
Take $x$ root of the inequality: $$(1-a)(1+an)^{1/a}\leqslant (1-n)(1+n)=1-n^2<1$$ But what if $a_1,\ldots,a_n$ do not have uniform values?
By AM-GM $$\prod_{i=1}^n(1-a_i)\left(1+\sum_{i=1}^na_i\right)<\left(\frac{\sum\limits_{i=1}^n(1-a_i)+1+\sum\limits_{i=1}^na_i}{n+1}\right)^{n+1}=\left(\frac{n+1}{n+1}\right)^{n+1}=1.$$ The equality does not occur because $1-a_i<1+\sum\limits_{i=1}^na_i.$