How to prove the Weierstrass product inequality using recurrency? I have to prove the following the Weierstrass Product Inequality  product inequality using recurrency .
$$ \forall n \in N \setminus\{0 , 1\}; 1 - \sum_{k=0}^n a_k < \prod (1-a_k) < \left(1 + \sum_{k=0}^n a_k\right)^{-1} $$
with $ 0 < a_k < 1 $
I tried to prove it for $n_0 = 2$ , we get :
$$ 1 - a_1 - a_2 < (1- a_1)\cdot(1-a_2)$$
I don't even find a way to prove this part .
I'm looking for some hint that would help me understand how I can prove this inequality using recurrency . are there some inequality theorem  could use ? 
My intuition is telling me that the fact : $ 0 < a_k < 1 $ is key to solving this, but I still can't find the path for it .
 A: The left inequality is a linear inequality of all $a_k$,
which says that it's enough to check it for $a_k\in\{0,1\}$, which is obviously true for $a_k=1$.
For $a_k=0$ it reduces to  $n=2$, which is
$$1-a_1-a_2<1-a_1-a_2+a_1a_2,$$ which is obvious. 
The right inequality we can prove by induction.
Indeed, the base it's $(1-a_1)(1+a_1)<1,$ which is obvious.
Let $$\prod_{k=1}^n(1-a_k)\left(1+\sum_{k=1}^na_k\right)<1.$$
Thus, $$\prod_{k=1}^{n+1}(1-a_k)\left(1+\sum_{k=1}^{n+1}a_k\right)=$$
$$=(1-a_{n+1})\prod_{k=1}^n(1-a_k)\left(1+\sum_{k=1}^na_k+a_{n+1}\right)<$$
$$<1-a_{n+1}+a_{n+1}(1-a_{n+1})\prod_{k=1}^n(1-a_k)<1$$
A: Alternative Proof (left inequality)
We prove that $\forall a_i \in [0,1], i=1, 2, \ldots, n,$
$$
\prod _ { i = 1 } ^ n (1- a _ i) + \sum _ { i = 1 } ^ n a _ i - 1 =\sum_{i=1}^{n-1}  a_i\left(1-\prod_{j>i} (1-a_j)\right)\geqslant 0.\tag1
$$
It suffices to prove just the equality above.
Denote $P_i=\prod_{j=i}^n (1-a_j)$. Note that $(1-a_i) P_{i+1}=P_i, \forall i <n$, and $1 - a_n =P_n$.
Then
$$\sum_{i=1}^{n-1} a_i\left(1-\prod_{j>i}(1- a_j)\right)=\sum_{i=1}^{n-1} a_i\left(1-P_{i+1}\right)\\
=\sum_{i=1}^{n-1} (a_i -P_{i+1}+P_i)=\sum_{i=1}^{n-1} a_i - \sum_{i=1}^{n-1} P_{i+1} + \sum_{i=1}^{n-1} P_i\\
=\sum_{i=1}^{n-1} a_i - \left(\sum_{i=2}^{n-1} P_i + P_n \right) + \left(P_1 + \sum_{i=2}^{n-1} P_i\right) = P_1 - P_n + \sum_{i=1}^{n-1} a_i \\
=\prod_{i=1}^n (1-a_i)-(1-a_n)+\sum_{i=1}^{n-1} a_i = \prod_{i=1}^n (1-a_i)+\sum_{i=1}^{n} a_i -1 .
$$
When do we have equality? WLOG assume $a_1 \leqslant a_2 \leqslant \cdots \leqslant a_n$, we must have $$a_{n-1}a_n=0 \implies a_{n-1}=0 \implies a_1=a_2 = \cdots =a_{n-1}=0.$$
On the other hand if $n-1$ of the $a_i's$ are zero, then the equality holds. Therefore $$"=" \iff (n-1) \text{ of } a_i's \text{ are equal to zero}.\tag2 $$
