How do I show that $\sum_{i = 1}^n \frac 1{\sqrt{a_n}} \lt \frac {\sqrt 3}6$ for $a_n = 4n(4n + 1)(4n + 2)$? Let $a_n = 4n(4n + 1)(4n + 2)$, show that $$\sum_{i = 1}^n \frac 1{\sqrt{a_i}} \lt \frac {\sqrt 3}6 \quad \forall n \in \mathbb{N}^+.$$

I know I need to find an upper bound for $1/\sqrt{a_n}$ but I can't see how, especially with the square root. Any hints will be appreciated!
We may notice that:
$$ \forall n\geq 1,\qquad  \frac{1}{\sqrt{4n(4n+1)(4n+2)}}\leq \frac{1}{2}\left(\frac{1}{\sqrt{4n-1}}-\frac{1}{\sqrt{4n+3}}\right) $$
hence it follows that:
$$ \sum_{n=1}^{N}\frac{1}{\sqrt{4n(4n+1)(4n+2)}} &lt; \left.\frac{1}{2\sqrt{4n-1}}\right|_{n=1}=\color{red}{\frac{1}{2\sqrt{3}}}$$
as wanted. Creative telescoping wins again.
you can write inequality as $2\sqrt{3}&lt;\sqrt{4n(4n+1)(4n+2)}$ squaring we get it as $12&lt;4n(4n+1)(4n+2)$ this the function is continuously increasing as $n$ is positive. Also $n\in N+$ this base case is $1$ so plugging in we get $120$ this $12&lt;120$ so it's true for akin as function is monotonic and increasing