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How to show

$x_1,x_2, \dots ,x_n \geq 0 $ and $ x_1 + x_2 + \dots + x_n \leq \frac{1}{2} \implies (1-x_1)(1-x_2) \cdots (1-x_n) \geq \frac{1}{2}$

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    Is it your homework ?2012-10-19
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    Is a problem that I found and I can't solve2012-10-19
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    I think we should start with induction.2012-10-19
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    @sebastian azocar:- Basically it is the Weierstrass Product Inequality2012-10-19
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    I've changed [tag:algebra] tag to [tag:algebra-precalculus], since we don't use algebra tag anymore, see [meta](http://meta.math.stackexchange.com/questions/473/the-use-of-the-algebra-tag/3081#3081) for details.2012-10-19
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    This is also related: [Proving $\prod \limits_{k=0}^{n}(1-a_k) \geq1- \sum\limits_{k=0}^{n}a_k$](http://math.stackexchange.com/q/616722).2017-01-06

2 Answers 2

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It is easy to see that:

$$(1-a)(1-b) \geq 1-(a+b)$$

Then, you can use induction to prove that:

$$(1-x_1)(1-x_2)...(1-x_n) \geq 1-(x_1+x_2+...+x_n)$$

The inductive step is:

$$(1-x_1)(1-x_2)...(1-x_n)(1-x_{n+1}) \geq \left[ 1-(x_1+x_2+...+x_n) \right] (1-x_{n+1}) \geq 1-(x_1+x_2+...+x_n+x_{n+1})$$

For this to work you only need that all $1-x_i \geq 0$...Of course you need $x_1+..+x_n \leq \frac{1}{2}$ to get the desired inequality.

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Hint: Looks like a good candidate for induction. The base case is easy, $n=1$ says $x_1 \le \frac 12 \implies 1-x_1 \ge \frac 12$ Intuitively, the limit on the sum of the $x_i$ says if you expand the product the second term is less than $\frac 12$, and the third is positive. Can you show that each positive term dominates the negative term that follows?

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    I tried, but is tricky, I think must exist a solution more easy2012-10-19
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    @sebastianazocar: and N. S. has it.2012-10-19