MHB Can Sum to Product Inequalities Hold for Non-Negative Reals?

[lfdahl]

Given non-negative reals, $\alpha_i$, where $i = 1,2,...,n.$

Prove, that

$\alpha_1+\alpha_2+...+\alpha_n \leq \frac{1}{2}$ $\Rightarrow$ $(1-\alpha_1)(1-\alpha_2)...(1-\alpha_n) \geq \frac{1}{2}.$

 

[MountEvariste]

Spoiler

Immediate consequence of Weiertrass inequality:

$$\prod_{1 \le k \le n}(1-a_k) \ge 1-\sum_{1 \le k \le n}a_k \ge 1-\frac{1}{2} = \frac{1}{2}.$$

There's an elegant proof of Weierstrass inequality.

 

[lfdahl]

Spoiler

Immediate consequence of Weiertrass inequality:

$$\prod_{1 \le k \le n}(1-a_k) \ge 1-\sum_{1 \le k \le n}a_k \ge 1-\frac{1}{2} = \frac{1}{2}.$$

There's an elegant proof of Weierstrass inequality.

Thankyou for a clever solution, June29, and for your participation!(Cool)

 

[lfdahl]

Can anyone prove the above statement by induction? (Wave)

 

[MountEvariste]

Can anyone prove the above statement by induction? (Wave)

Spoiler

It's obviously true for $n=1$ since we have $ {\alpha}_1 \leqslant \frac{1}{2} = 1-\frac{1}{2} \implies 1-\alpha_1 \geqslant \frac{1}{2}. $

Now, suppose it's true for $n = k \in \mathbb{N}$. We shall prove that it's true for $n=k+1$. $\displaystyle \begin{aligned} \frac{1}{2} & \leqslant 1- \sum_{1 \leqslant j \leqslant k+1}\alpha_j = 1-\alpha_{k+1}-\sum_{1 \leqslant j \leqslant k}\alpha_j \leqslant 1-\alpha_{k+1}-\sum_{1 \leqslant j \leqslant k}\alpha_j+a_{k+1}\sum_{1 \leqslant j \leqslant k} \alpha_j \\& =\left(1-\alpha_{k+1}\right)\left(1-\sum_{1 \leqslant j \leqslant k} \alpha_j\right)
\leqslant \left(1-\alpha_{k+1}\right) \prod_{1 \leqslant j \leqslant k} \left(1-\alpha_j \right) = \prod_{1 \leqslant j \leqslant k+1} \left(1-\alpha_j \right) \end{aligned} $

So it's true for $n=k+1$. Since it's true for $n=1, k+1$, it's true for all $n\in\mathbb{N}$.

 

[lfdahl]

Spoiler

It's obviously true for $n=1$ since we have $ {\alpha}_1 \leqslant \frac{1}{2} = 1-\frac{1}{2} \implies 1-\alpha_1 \geqslant \frac{1}{2}. $

Now, suppose it's true for $n = k \in \mathbb{N}$. We shall prove that it's true for $n=k+1$. $\displaystyle \begin{aligned} \frac{1}{2} & \leqslant 1- \sum_{1 \leqslant j \leqslant k+1}\alpha_j = 1-\alpha_{k+1}-\sum_{1 \leqslant j \leqslant k}\alpha_j \leqslant 1-\alpha_{k+1}-\sum_{1 \leqslant j \leqslant k}\alpha_j+a_{k+1}\sum_{1 \leqslant j \leqslant k} \alpha_j \\& =\left(1-\alpha_{k+1}\right)\left(1-\sum_{1 \leqslant j \leqslant k} \alpha_j\right)
\leqslant \left(1-\alpha_{k+1}\right) \prod_{1 \leqslant j \leqslant k} \left(1-\alpha_j \right) = \prod_{1 \leqslant j \leqslant k+1} \left(1-\alpha_j \right) \end{aligned} $

So it's true for $n=k+1$. Since it's true for $n=1, k+1$, it's true for all $n\in\mathbb{N}$.

A nice solution, June29! Thankyou for your participation!

Spoiler

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