Let \(y = \frac{x^2 - x + 1}{x^2 + x + 1}\). Then \((y - 1)x^2 + (y + 1)x + (y - 1) = 0\).

For real x, the discriminant must be ≥ 0: \((y+1)^2 - 4(y-1)^2 \ge 0 \Rightarrow (3 - y)(3y - 1) \ge 0 \Rightarrow \frac{1}{3} \le y \le 3\).

Minimum = \(\frac{1}{3}\) (at x = 1).

Answer: (D) \(\frac{1}{3}\)