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 $1^2 + 2^2 + ..... + (n-1)^2 < \frac {n^3} { 3} < 1^2 + 2^2 + ...... + n^2$.
$f(x)$ is an increasing function, so left Riemann sums under-approximate, and right Riemann sums over-approximate. Using a step size of 1, the inequality is $$\sum_{i=1}^n \sqrt{i-1} < \int_0^n \sqrt{x} \, \mathrm{d}x < \sum_{i=1}^{n} \sqrt{i}.$$
by induction :
for LHS: if $n=k ,\sum _{i=1}^k \sqrt{i-1} \le \dfrac{2k\sqrt{k}}{3}$
when $n=k+1$, we have to prove $\dfrac{2k\sqrt{k}}{3}+\sqrt{k} \le \dfrac{2(k+1)\sqrt{k+1}}{3} \iff 3k\ge 4$ which is true when $k\ge2$
similar method to RHS.
$\sqrt{x}$ 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.
