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 .
Best Answer
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, $$\prod_{k=1}^{n+1}(1-a_k)\left(1+\sum_{k=1}^{n+1}a_k\right)=$$ $$=(1-a_{n+1})\prod_{k=1}^n(1-a_k)\left(1+\sum_{k=1}^na_k+a_{n+1}\right)<$$ $$<1-a_{n+1}+a_{n+1}(1-a_{n+1})\prod_{k=1}^n(1-a_k)<1$$