Establishing Weierstrass inequalities 
Suppose $0< a_k < 1$ $k=1,2,...,n$ and $a_1+...+a_n < 1 $. Prove that
${\bf A.}$ $ 1 + \sum_{k=1}^n a_k < \prod_{k=1}^n (1+a_k) <
 \dfrac{1}{1-\sum a_k } $
${\bf B.}$ $ 1 - \sum_{k=1}^n a_k < \prod_{k=1}^n (1-a_k) <
 \dfrac{1}{1+\sum a_k } $

attempt.
Equivalently we can show that
$$ (1 + \sum a_k ) (1- \sum a_k) < \prod(1+a_k) \cdot (1-\sum a_k) < 1 $$
or
$$ 1 - ( \sum a_k )^2 < \prod (1+a_k) - \prod (1+a_k) \sum a_k < 1 $$
Notice that $( \sum a_k)^2 $ is a number less than one so $1 - (\sum a_k)^2 $ is less than one. I am having some difficulty trying to establish rtaht $\prod(1+a_k) - \prod(1+a_k) \sum a_k < 1$.
For the other part of the inequality, we can let $x = \prod (1+a_k)$ where $0<x<2$ and $y = \sum a_k$ where $0<y<1$ and so if we prove
$$ 1 - y^2 < x - xy $$
then wed have the result. Equivalently, we can write $\dfrac{1-y^2}{1-y} < x  \implies 1+y<x$ but this may not always be true. Any hint?
 A: For the 1st part, notice that $(1+x)(1+y)=1+x+y+xy > 1+x+y$. Then,
$$\prod_{k=1}^n (1+a_k) > (1+a_1+a_2)\prod_{k=3}^n(1+a_k)>(1+a_1+a_2+a_3)\prod_{k=4}^n(1+a_k)>....>1 + \sum_{k=1}^n a_k$$
The second inequality is just the application of the AM-GM inequality for $n+1$ positive numbers
 $$(1-\sum a_k)\prod_{k=1}^n (1+a_k) \leq (\frac{1-\sum a_k + \sum (1+a_k)}{n+1})^{n+1}=1$$
Equality happens when $a_1=a_2=...=a_k=\sum a_k$ which is impossible. So we have strict inequality here.
For the 2nd part, Using AM-GM and the same trick with $(1-x)(1-y)=1-x-y+xy > 1-x-y$. 
A: The first part can be proved by induction with the base case $1+a_1 < \frac{1}{1-a_1}$ being trivial. Assume $\prod_{k=1}^{n-1} (1+a_k) < \frac{1}{1-\sum_{k=1}^{n-1}a_k}$. 
Multiplying both sides by $(1+a_n)$, it suffices to show that $\frac{1+a_n}{1-\sum_{k=1}^{n-1}a_k} < \frac{1}{1-\sum_{k=1}^{n}a_k}$ which is an easy computation.
For the second part, it is indeed true that $1 + y  < x$: When expanding the product $x$, we get the constant term $1$, the linear terms $\sum a_k$ along with the other terms $\sum a_ia_j$, etc. The constant and linear terms give $1+y$, and the other terms are positive. This can also be proved more formally by induction.
