We wish to show that $\frac{1}{(n^2+n)^\frac{3}{4}} > \frac{2}{\sqrt{n}} - \frac{2}{\sqrt{n+1}}$

Thus, \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*}

easy bash $\frac{1}{(n^2+n)^\frac{3}{4}} > \frac{2}{\sqrt{n}} - \frac{2}{\sqrt{n+1}}$ $\frac{1}{(n^2+n)^\frac{3}{4}} > 2 \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n^2+n}}$ $\frac{1}{(n^2+n)^\frac{1}{4}}>2 ({\sqrt{n+1}-\sqrt{n}})$ $\frac{1}{(n^2+n)^\frac{1}{2}}> 4(2n+1-2\sqrt{n^2+n})$ $4(2n+1)< \frac{1}{(n^2+n)^\frac{1}{2}}+8\sqrt{n^2+n}$ $16(4n^2+4n+1)<64n^2+64n+16+\frac{1}{(n^2+n)^\frac{1}{4}}$ - which is true

ACGNmath wrote: We wish to show that $\frac{1}{(n^2+n)^\frac{3}{4}} > \frac{2}{\sqrt{n}} - \frac{2}{\sqrt{n+1}}$

Thus, \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*} Easier is $\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}}} $

cjquines0 wrote: Let \(n\) be any positive integer. Prove that \[\sum^n_{i=1} \frac{1}{(i^2+i)^{3/4}} > 2-\frac{2}{\sqrt{n+1}}\]. Stronger \[\sum^n_{i=1} \frac{1}{(i^2+i)^{3/4}} \ge \frac{2}{\sqrt{1-a}}-\frac{2}{\sqrt{n+1-a}}\]Here $\sqrt{(1-a)(2-a)}(\sqrt{1-a}+\sqrt{2-a})=\sqrt[4]{8}.$