Prove that for any $k \mathbb \in N^*$

Question

Prove that for any $k \mathbb \in N^*$: $$\frac{1}{2\sqrt{(k+1)^3}} \leq \frac{1}{\sqrt{k}}-\frac{1}{\sqrt {k+1}} \leq \frac{1}{2\sqrt{k^3}}$$

I have tried to use simple induction but i didn't get a good result

Answer 1

Notice that $$ \frac1{\sqrt{k}}-\frac1{\sqrt{k+1}}=\frac1{\sqrt{k}\sqrt{k+1}(\sqrt{k}+\sqrt{k+1})} $$ and we have $$ \sqrt{k}\sqrt{k+1}(\sqrt{k}+\sqrt{k+1})\le \sqrt{k+1}\sqrt{k+1}(\sqrt{k+1}+\sqrt{k+1}) = 2\sqrt{(k+1)^3}$$ and similarly $$ \sqrt{k}\sqrt{k+1}(\sqrt{k}+\sqrt{k+1})\ge \sqrt{k}\sqrt{k}(\sqrt{k}+\sqrt{k}) = 2\sqrt{k^3}$$ and your inequality follows.