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?
 A: 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,
$$
\sum_{i = 1}^n \frac{1}{(i^2 + i)^{\frac34}} > 2\left(1 - \frac{1}{\sqrt{n+1}}\right).
$$
A: I think it is a good idea to prove the given inequality with First Principle of Mathematical  Induction.
1) Step-1:
To see if this inequality is true for $n=1$,
$$ \frac{1}{2^ \frac{3}{4}}>2- \sqrt{2}$$
Which is true (substitute $2^{ \frac{1}{4}}=k$ where and $k>1$ you will get result). Hence $P(1)$ is true.
2) Step-2
Assuming that $P(k-1)$ is true we'll prove that if $P(k)$ will be true.
$$\sum_{i=1}^{k-1} \frac{1}{(i²+i)^\frac{3}{4}}>2-\frac{2}{\sqrt{k}}$$
Adding $\frac{1}{({k²+k})^\frac{3}{4}}$ both sides,
$$\sum_{i=1}^{k} \frac{1}{(i²+i)^\frac{3}{4}}>2-\frac{2}{\sqrt{k}}+\frac{1}{{(k²+k})^\frac{3}{4}}$$
Now we have to prove
$$2-\frac{2}{\sqrt{k}}+\frac{1}{{(k²+k})^\frac{3}{4}}>2-\frac{2}{\sqrt{k+1}}$$
Or,
$$\frac{2(\sqrt{k+1}-\sqrt{k})}{\sqrt{k(k+1)}}<\frac{1}{{(k(k+1))^\frac{3}{4}}}$$
Note that $1=(\sqrt{k+1})^2-(\sqrt{k})^2$, After substitution at RHS numerator and cancelling we'll get,$$2<\frac{\sqrt{k+1}+\sqrt{k}}{(k(k+1))^{1/4}}$$
Or,
$$2< \frac{(k+1)^{\frac{1}{4}}}{k^{\frac{1}{4}}}+\frac{k^{\frac{1}{4}}}{(k+1)^{\frac{1}{4}}}$$
Which is a trivial result of A.M.$\geq$ G.M.
Hence we're done of second step, and also the problem.
Therefore our claim is true!
A: 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*}
