1. **State the problem:** Solve the inequality $$|x^2 - x + 1| \geq 1$$. 2. **Recall the definition of absolute value inequality:** For any expression $A$, $$|A| \geq 1$$ means $$A \geq 1$$ or $$A \leq -1$$. 3. **Apply this to our problem:** $$x^2 - x + 1 \geq 1$$ or $$x^2 - x + 1 \leq -1$$. 4. **Solve each inequality separately:** - For $$x^2 - x + 1 \geq 1$$: $$x^2 - x + 1 - 1 \geq 0 \implies x^2 - x \geq 0$$. Factor: $$x(x - 1) \geq 0$$. This product is nonnegative when both factors are nonnegative or both are nonpositive. - Case 1: $$x \geq 0$$ and $$x - 1 \geq 0 \implies x \geq 1$$. - Case 2: $$x \leq 0$$ and $$x - 1 \leq 0 \implies x \leq 0$$. So, solution for this part is $$x \leq 0$$ or $$x \geq 1$$. - For $$x^2 - x + 1 \leq -1$$: $$x^2 - x + 1 + 1 \leq 0 \implies x^2 - x + 2 \leq 0$$. 5. **Analyze the quadratic $$x^2 - x + 2$$:** Calculate discriminant: $$\Delta = (-1)^2 - 4 \times 1 \times 2 = 1 - 8 = -7 < 0$$. Since $$\Delta < 0$$ and leading coefficient is positive, $$x^2 - x + 2 > 0$$ for all real $$x$$. Therefore, $$x^2 - x + 2 \leq 0$$ has no real solutions. 6. **Combine the solutions:** The solution to the original inequality is $$x \leq 0$$ or $$x \geq 1$$. **Final answer:** $$\boxed{(-\infty, 0] \cup [1, \infty)}$$