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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 .

Michael Rozenberg
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Anis Souames
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2 Answers2

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

Michael Rozenberg
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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 $$"=" \iff (n-1) \text{ of } a_i's \text{ are equal to zero}.\tag2 $$

Neat Math
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