<s>[INST] Prove that $\sqrt{n+1}>\sqrt{n}+\frac{1}{2\sqrt{n}}-\frac{1}{8n\sqrt{n}}$ 
Prove that $\sqrt{n+1}>\sqrt{n}+\frac{1}{2\sqrt{n}}-\frac{1}{8n\sqrt{n}}$ if $n>0$. 

I didn't see an easy way of proving this without doing a lot of algebra and rearranging. Is there an easier way?
 [/INST] It does not look so terrible to me! We have:
$$ \sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}} \tag{1}$$
and:
$$ \frac{1}{2\sqrt{n}}-\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n}\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{1}{2\sqrt{n}\left(\sqrt{n}+\sqrt{n+1}\right)^2}\tag{2}$$
so the inequality boils down to $\sqrt{n}+\sqrt{n+1}>2\sqrt{n}$, quite trivial.
</s>