<s>[INST] 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?
 [/INST] Note that since
$$
(i^2 + i)^{\frac34} = (i^2 + i)^{\frac12} \cdot i^\frac14 \cdot (i+1)^\frac14 < \sqrt{i} \cdot \sqrt{i+1} \cdot \frac{\sqrt{i} + \sqrt{i+1} }{2} = \frac12 \cdot \frac{\sqrt{i} \cdot \sqrt{i+1}}{\sqrt{i+1} - \sqrt{i}},
$$
we have
$$
\frac{1}{(i^2 + i)^{\frac34}} > \frac{2(\sqrt{i+1} - \sqrt{i})}{\sqrt{i}\cdot\sqrt{i+1} \cdot} = 2 \left(\frac{1}{\sqrt{i}} - \frac{1}{\sqrt{i+1}}\right).
$$
Therefore,
$$
\sum_{i = 1}^n \frac{1}{(i^2 + i)^{\frac34}} > 2\left(1 - \frac{1}{\sqrt{n+1}}\right).
$$
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