<s>[INST] Prove that the sequence given by its general term $\frac{\sqrt{n^2+1}-n}{\sqrt{n+3}+2n}$ is strictly decreasing. Prove that the sequence given by its general term: 
$$\frac{\sqrt{n^2+1}-n}{\sqrt{n+3}+2n}$$
is strictly decreasing. 
$a_n>a_{n+1} \Leftrightarrow \frac{\sqrt{n^2+1}-n}{\sqrt{n+3}+2n}>\frac{\sqrt{(n+1)^2+1}-(n+1)}{\sqrt{(n+1)+3}+2(n+1)}$
Is there some kind of trick to apply in these sorts of problems or do I have to solve this by sheer algebraic manipulation? If I subtract $\frac{\sqrt{(n+1)^2+1}-(n+1)}{\sqrt{(n+1)+3}+2(n+1)}$ from both sides I'll end up with something really ugly so I'm wondering if there's a better way. 
 [/INST] Hint:
$$a_n=\frac{\sqrt{n^2+1}-n}{\sqrt{n+3}+2n}=\frac{(\sqrt{n^2+1}-n)(\sqrt{n^2+1}+n)}{(\sqrt{n+3}+2n)(\sqrt{n^2+1}+n)}=\frac1{(\sqrt{n+3}+2n)(\sqrt{n^2+1}+n)}$$
</s>