Attempting to prove the following inequality

Question

I got a question on a test to prove the following inequality:

$$ \sqrt{1*2}\space+\space\sqrt{2*3}\space+\space...\space+\space\sqrt{n(n-1)}>\frac{n^2-1}{2} : n>1, n\in\mathbb{N} $$ I tried to prove this with induction and got to the point where I proved

$$ n=k+1 \\ \sqrt{1*2}\space+\space\sqrt{2*3}\space+\space...\space+\space\sqrt{k(k+1)}>\frac{k^2-1}{2}\space+\sqrt{k(k+1)} \\ $$ and I also proved $$ \frac{(k+1)^2-1}{2} > \frac{k^2-1}{2}\space+\sqrt{k(k+1)} $$ but I couldn't successfully prove $$ \sqrt{1*2}\space+\space\sqrt{2*3}\space+\space...\space+\space\sqrt{k(k+1)}>\frac{(k+1)^2-1}{2} $$

Answer 1

I can prove the opposite inequality.

According to GM-AM you have for $n > 1$: \begin{eqnarray*} \color{blue}{\sum_{i=1}^{n-1}\sqrt{i(i+1)}} & \color{blue}{<} & \sum_{i=1}^{n-1}\frac{i + (i+1)}{2} \\ & = & \frac{1}{2}\left(\sum_{i=1}^{n-1}2i + \sum_{i=1}^{n-1} 1 \right)\\ & = & \frac{1}{2}\left(n(n-1) + (n-1)\right)\\ & = & \color{blue}{\frac{n^2-1}{2}}\\ \end{eqnarray*}

Edit after comments:

If you add the next term $\sqrt{n(n+1)}$, then the relation sign changes its direction:

• $\sum_{i=1}^{\color{red}{n}}\sqrt{i(i+1)} \color{red}{>} \frac{n^2-1}{2}$

The proof for this case you can see in Michael Rosenberg's answer.

Answer 2

By C-S we can get the following estimation.

$$\sum_{k=1}^{n-1}\sqrt{k(k+1)}=\sum_{k=1}^{n-1}\frac{k^2+k}{\sqrt{k(k+1)}}\geq\sum_{k=1}^{n-1}\frac{k^2+\frac{k}{2}+\frac{k}{2}}{k+\frac{1}{2}}=\sum_{k=1}^{n-1}k+\sum_{k=1}^{n-1}\frac{k}{2k+1}=$$ $$=\frac{n^2-n}{2}+\sum_{k=1}^{n-1}\frac{k^2}{2k^2+k}\geq\frac{n^2-n}{2}+\frac{\left(\sum\limits_{k=1}^{n-1}k\right)^2}{\sum\limits_{k=1}^{n-1}(2k^2+k)}=$$ $$=\frac{n^2-n}{2}+\frac{\left(\frac{n(n-1)}{2}\right)^2}{\frac{n(n-1)(2n-1)}{3}+\frac{n(n-1)}{2}}=\frac{n^2-1}{2}\cdot\frac{4n}{4n+1}.$$