1. **State the problem:** Solve the inequality $1 \leq n - 4\sqrt{n}$ for $n \geq 0$ since $\sqrt{n}$ is defined only for non-negative $n$. 2. **Rewrite the inequality:** $$1 \leq n - 4\sqrt{n}$$ 3. **Substitute:** Let $x = \sqrt{n}$, so $n = x^2$ and $x \geq 0$. 4. **Rewrite in terms of $x$:** $$1 \leq x^2 - 4x$$ 5. **Bring all terms to one side:** $$x^2 - 4x - 1 \geq 0$$ 6. **Solve the quadratic inequality:** The quadratic is $x^2 - 4x - 1 = 0$. 7. **Find roots using quadratic formula:** $$x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-1)}}{2} = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$$ 8. **Determine intervals:** Since the parabola opens upward (coefficient of $x^2$ is positive), the inequality $x^2 - 4x - 1 \geq 0$ holds for: $$x \leq 2 - \sqrt{5} \quad \text{or} \quad x \geq 2 + \sqrt{5}$$ 9. **Consider domain $x \geq 0$:** Note that $2 - \sqrt{5} \approx 2 - 2.236 = -0.236$ which is negative, so discard this root for $x \geq 0$. 10. **Final solution for $x$:** $$x \geq 2 + \sqrt{5}$$ 11. **Back-substitute $x = \sqrt{n}$:** $$\sqrt{n} \geq 2 + \sqrt{5}$$ 12. **Square both sides:** $$n \geq (2 + \sqrt{5})^2 = 4 + 4\sqrt{5} + 5 = 9 + 4\sqrt{5}$$ 13. **Answer:** The solution to the inequality is: $$n \geq 9 + 4\sqrt{5}$$