<s>[INST] A simple double inequality with roots Prove or disprove for integer $n$ greater than $1$: 
$$\sqrt{1} + \sqrt{2} + ...+ \sqrt{n-1} < \frac{2n\sqrt{n}}{3} < \sqrt{1} + \sqrt{2} + ...+ \sqrt{n} $$
I think I know a solution, but I am looking for different approaches.
EDIT: Inspired by Prove that $1^2 + 2^2 + ..... + (n-1)^2 < \frac {n^3} { 3} < 1^2 + 2^2 + ...... + n^2$.
 [/INST] $\sqrt{x}$ is a concave increasing function on $\mathbb{R}^+$, hence by the Hermite-Hadamard inequality:
$$\frac{1}{2}\cdot\frac{1}{1}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n-1}}+\frac{1}{2\sqrt{n}}\leq \int_{1}^{n}\sqrt{x}\,dx = \frac{2}{3}\left(n\sqrt{n}-1\right)\tag{1}$$
as well as:
$$ \frac{(2n+1)\sqrt{2n+1}-1}{3\sqrt{2}}=\int_{1/2}^{n+1/2}\sqrt{x}\,dx\leq\frac{1}{\sqrt{1}}+\ldots+\frac{1}{\sqrt{n}}\tag{2}$$
improving a bit your original inequality.
</s>