How many $x$ ?

For fact, we know that a Arithmetic Mean $\geq$ Geometric Mean of the same terms. So:

$\frac{\frac{1}{x^{1000}} + \frac{1}{x^{999}} + ... + \frac{1}{x} + 1 + x + x^2 + ... + x^{999} + x^{1000}}{2001} \geq \sqrt[2001]{ \frac{1}{x^{1000}} \times \frac{1}{x^{999}} \times ... \times 1 \times x \times x^2 \times ... \times x^{999} \times x^{1000} }$

Which give us:

$\frac{1}{x^{1000}} + \frac{1}{x^{999}} + ... + \frac{1}{x} + 1 + x + x^2 + ... + x^{999} + x^{1000} \geq 2001 \sqrt[2001]{1}$ $\therefore \frac{1}{x^{1000}} + \frac{1}{x^{999}} + ... + \frac{1}{x} + 1 + x + x^2 + ... + x^{999} + x^{1000} \geq 2001$

Anything wrong, give me a feedback.

Since we have $x^i$ to all be positive for integers $-1000 \leq i \leq 1000$ , we can apply the Cauchy-Schwarz inequality to get:

$(\frac{1}{x^{1000}} + \frac{1}{x^{999}} + \ldots + \frac{1}{x} + 1 + x + x^2 +\ldots + x^{999} + x^{1000})^2$

$= ({x^{1000}} + {x^{999}} + \ldots + x + 1 + \frac{1}{x} + \frac{1}{x^2} +\ldots + \frac{1}{x^{999}} + \frac{1}{x^{1000}})(\frac{1}{x^{1000}} + \frac{1}{x^{999}} + \ldots + \frac{1}{x} + 1 + x + x^2 +\ldots + x^{999} + x^{1000})$

$\geq (\sqrt{x^{1000}\cdot \frac{1}{x^{1000}}} + \sqrt{x^{999}\cdot \frac{1}{x^{999}}} +\ldots + \sqrt{ \frac{1}{x^{999}}\cdot x^{999}} + \sqrt{ \frac{1}{x^{1000}}\cdot x^{1000}})^2$

$=2001^2$

With equality when $\frac{1}{x^{1000}} = \frac{1}{x^{999}} = \ldots = \frac{1}{x} = 1 = x = x^2 =\ldots = x^{999} = x^{1000}$ , or when $x=1$