1. **State the problem:** Solve the inequality $4n^{\frac{1}{2}} + n \leq 2n$ for $n$. 2. **Rewrite the inequality:** $$4\sqrt{n} + n \leq 2n$$ 3. **Bring all terms to one side:** $$4\sqrt{n} + n - 2n \leq 0$$ $$4\sqrt{n} - n \leq 0$$ 4. **Rewrite as:** $$4\sqrt{n} \leq n$$ 5. **Since $n \geq 0$ (because of the square root), divide both sides by $\sqrt{n}$ (which is positive for $n>0$):** $$4 \leq \frac{n}{\sqrt{n}} = \sqrt{n}$$ 6. **So:** $$\sqrt{n} \geq 4$$ 7. **Square both sides:** $$n \geq 16$$ 8. **Check $n=0$:** At $n=0$, $4\sqrt{0} + 0 = 0 \leq 0$ is true. 9. **Check values between 0 and 16:** For example, at $n=1$, $4(1) + 1 = 5 \leq 2(1) = 2$ is false. 10. **Conclusion:** The solution is $n=0$ or $n \geq 16$. **Final answer:** $$n=0 \quad \text{or} \quad n \geq 16$$