\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:
$$
\frac{a_{n+1}}{a_n} = \frac{\sqrt{n+5}+\sqrt{n}}{\sqrt{n+6}+\sqrt{n+1}} < 1
$$