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

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 ? – Nikita Evseev Oct 19 '12 at 4:10
Is a problem that I found and I can't solve – sebastian azocar Oct 19 '12 at 4:20
I think we should start with induction. – Nikita Evseev Oct 19 '12 at 4:23
@sebastian azocar:- Basically it is the Weierstrass Product Inequality – Souvik Dey Oct 19 '12 at 4:32
I've changed algebra tag to algebra-precalculus, since we don't use algebra tag anymore, see meta for details. – Martin Sleziak Oct 19 '12 at 6:51

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 easy – sebastian azocar Oct 19 '12 at 4:28
@sebastianazocar: and N. S. has it. – Ross Millikan Oct 19 '12 at 4:34