$\sqrt{k}+\frac{N}{k}\geq N^{1/3}$ where $N, k\geq 1$.

Question

$$\sqrt{k}+\frac{N}{k}\geq N^{1/3}$$

I did the case where $N=k$ and showed that one function approaches infinity faster than the other. But how do I do it in the other cases?

Answer 1

For positives $k$ and $N$ by AM-GM $$\sqrt{k}+\frac{N}{k}=2\cdot\frac{\sqrt{k}}{2}+\frac{N}{k}\geq3\sqrt[3]{\left(\frac{\sqrt{k}}{2}\right)^2\frac{N}{k}}=\frac{3}{\sqrt[3]4}\sqrt[3]{N}>\sqrt[3]{N}.$$

Answer 2

An alternative:

For $k \le N^{2/3}$:

$$\sqrt k + \frac Nk\ge \sqrt k+ N^{1/3}> N^{1/3}$$

For $k\ge N^{2/3}$:

$$\sqrt {k}+\frac Nk \ge N^{1/3}+\frac Nk > N^{1/3}$$

Answer 3

$$\frac{k\sqrt{k} + N}{k}\geq N^{1/3}$$ $$k\sqrt{k} + N \geq kN^{1/3}$$ $$\frac{k\sqrt{k}}{N^{1/3}} + N^{2/3} \geq k$$ We leave proven the cases when the leftmost term can overcome $k$ on its own, and study when that doesn't happen $$\frac{k\sqrt{k}}{N^{1/3}} < k\implies N^{1/3}>\sqrt{k} \implies N^{2/3}>k$$ and so the right term is able to overcome $k$ in the other case, QED