<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] 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!
</s>