Prove $\sum_{i=1}^{n} \frac{1}{(i²+i)^\frac{3}{4}}>2- \frac{2}{\sqrt{n+1}}$ 
Let $i\in\mathbb{N}$ $$\sum_{i=1}^{n} \frac{1}{(i²+i)^\frac{3}{4}}>2-\frac{2}{\sqrt{n+1}}$$

A.M.-G.M. inequality makes this problem complex to solve.
It was true for $n=1,2,3,4$ then solving this became difficult. How can I solve this?

We first observe that
$$\frac{2}{\sqrt{n}} - \frac{2}{\sqrt{n+1}} = \frac{2}{\sqrt{n(n+1)}(\sqrt{n} + \sqrt{n+1})} 
< \frac{2}{\sqrt{n(n+1)}\cdot 2[n(n+1)]^{\frac{1}{4}}} = \frac{1}{(n^2+n)^{\frac{3}{4}}}$$
and then use this to argue
\begin{align*}
\displaystyle\sum_{i=1}^{n} \frac{1}{(i^2+i)^\frac{3}{4}} > \displaystyle\sum_{i=1}^{n} \frac{2}{\sqrt{i}} - \frac{2}{\sqrt{i+1}} = 2-\frac{2}{\sqrt{n+1}}
\end{align*}
