\begin{align*} a_n &= \sqrt{n+5}-\sqrt{n} = \left(\sqrt{n+5}-\sqrt{n}\right)\frac{\sqrt{n+5}+\sqrt{n}}{\sqrt{n+5}+\sqrt{n}} \\ &= \frac{\left(\sqrt{n+5}-\sqrt{n}\right)\left(\sqrt{n+5}+\sqrt{n}\right)}{\sqrt{n+5}+\sqrt{n}} \\ &= \frac{n+5 - n}{\sqrt{n+5}+\sqrt{n}} \\ &= \frac{5}{\sqrt{n+5}+\sqrt{n}} \\ \end{align*}
This is clearly a decreasing sequence. As $n$ increases, the denominator increases so the sequence as a whole decreases.
If you insist on using $\frac{a_{n+1}}{a_n}$, use the same approach above to show that:
\begin{align*} \frac{a_{n+1}}{a_n} &= \frac{\sqrt{n+6}-\sqrt{n+1}}{\sqrt{n+5}-\sqrt{n}} \\ &= \frac{\sqrt{n+5}+\sqrt{n}}{\sqrt{n+6}+\sqrt{n+1}} \end{align*}
Now, notice that:
$ n+1 \gt n \Rightarrow \sqrt{n+1} \gt \sqrt{n} $
And: $ n+6 \gt n+5 \Rightarrow \sqrt{n+6} \gt \sqrt{n+5} $
Add the inequalities side by side to get:
$ \sqrt{n+6} + \sqrt{n+1} \gt \sqrt{n+5} + \sqrt{n} $
Divide both sides by $\sqrt{n+6} + \sqrt{n+1}$ to get:
$ \frac{a_{n+1}}{a_n} = \frac{\sqrt{n+5}+\sqrt{n}}{\sqrt{n+6}+\sqrt{n+1}} < 1 $