Problem 1507

1. Initial Assumptions and Base Cases:
- We start by considering the initial value $x_1$. We analyze three cases:
- $x_1 < 0$
- $x_1 > 1$
- $x_1 \in [0, 1]$
- For $x_1 < 0$, $x_n > 1$ for $n > 1$, so $x_{2011} > \frac{2011}{2012}$ is true.
- For $x_1 > 1$, $x_2 < 0$ and $x_n > 1$ for $n > 2$, so $x_{2011} > \frac{2011}{2012}$ is true.
- We now focus on the case $x_1 \in [0, 1]$.

2. Induction Hypothesis:
- We will use mathematical induction to show that for $n \geq 3$, $x_n \in \left(1 - \frac{1}{n+1}, 1\right]$.
- Base Case:
- For $n = 3$, we have $x_3 = 1 - x_1 x_2$.
- Since $x_1 \in [0, 1]$ and $x_2 = 1 - x_1$, we have $x_1 x_2 = x_1 (1 - x_1) \leq \frac{1}{4}$ (since the maximum value of $x_1 (1 - x_1)$ is $\frac{1}{4}$ when $x_1 = \frac{1}{2}$).
- Therefore, $x_3 = 1 - x_1 (1 - x_1) \geq 1 - \frac{1}{4} = \frac{3}{4}$.
- Hence, $x_3 \in \left(1 - \frac{1}{4}, 1\right]$.

3. Inductive Step:
- Assume that for some $n \geq 3$, $x_n \in \left(1 - \frac{1}{n+1}, 1\right]$.
- We need to show that $x_{n+1} \in \left(1 - \frac{1}{n+2}, 1\right]$.
- From the recurrence relation, $x_{n+1} = 1 - x_1 x_2 \cdots x_n$.
- By the induction hypothesis, $x_n \in \left(1 - \frac{1}{n+1}, 1\right]$, so $x_1 x_2 \cdots x_n \in \left[0, \frac{1}{n+1}\right]$.
- Therefore, $1 - x_1 x_2 \cdots x_n \geq 1 - \frac{1}{n+1}$.
- We need to show that $1 - x_1 x_2 \cdots x_n < 1 - \frac{1}{n+2}$.
- Since $x_1 x_2 \cdots x_n \leq \frac{1}{n+1}$, we have:
$$1 - x_1 x_2 \cdots x_n \geq 1 - \frac{1}{n+1}$$
$$x_{n+1} \geq 1 - \frac{1}{n+1}$$
- To show the upper bound, we note that $x_1 x_2 \cdots x_n \leq \frac{1}{n+1}$, so:
$$x_{n+1} = 1 - x_1 x_2 \cdots x_n > 1 - \frac{1}{n+2}$$
- Hence, $x_{n+1} \in \left(1 - \frac{1}{n+2}, 1\right]$.

4. Conclusion:
- By induction, for $n \geq 3$, $x_n \in \left(1 - \frac{1}{n+1}, 1\right]$.
- Specifically, for $n = 2011$, we have $x_{2011} \in \left(1 - \frac{1}{2012}, 1\right]$.
- Therefore, $x_{2011} > \frac{2011}{2012}$.

The final answer is $\boxed{ x_{2011} > \frac{2011}{2012} }$.