understand the proof of $\frac{2 n}{3} \sqrt{n}<\sum_{k=1}^{n} \sqrt{k}<\frac{4 n+3}{6} \sqrt{n}$ 
If $n \in \mathbb{N}^*$,  prove that \begin{align*}\frac{2 n}{3} \sqrt{n}<\sum_{k=1}^{n} \sqrt{k}<\frac{4 n+3}{6} \sqrt{n}.\end{align*}

I am having trouble understanding the following proof of this problem:

Let
$a_{n}=(n+\lambda) \sqrt{n}, n \in \mathbb{N}^{*}$,
then
\begin{align} 
\frac{\sqrt{n}}{a_{n}-a_{n-1}} &=\frac{\sqrt{n}}{(n+\lambda) \sqrt{n}-(n-1+\lambda) \cdot \sqrt{n-1}} \\
&=\frac{\sqrt{n}}{(\sqrt{n}-\sqrt{n-1})(2 n-1+\sqrt{n(n-1)})+\lambda(\sqrt{n}-\sqrt{n-1})} \\
&=\frac{n+\sqrt{n(n-1)}}{2 n+\sqrt{n(n-1)}-1+\lambda} \\
&=\frac{2}{3}+\frac{\sqrt{n(n-1)}-n+2(1-\lambda)}{6 n+3 \sqrt{n(n-1)}-3(1-\lambda)} \tag{1}
\end{align}
for $n \in \mathbb{N}^*$, we have
$$\frac{1}{2}<n-\sqrt{n(n-1)} \leqslant 1 .$$
Substitute   $\lambda=\frac{1}{2}, \frac{3}{4}$  into the formula (1), respectively,  then we obtain that
$$ \tag{2}
\frac{2}{3}\left(n+\frac{1}{2}\right) \sqrt{n} \leqslant \sum_{k=1}^{n} \sqrt{k}<\frac{2}{3}\left(n+\frac{3}{4}\right) \sqrt{n}.
$$

Can anyone explain to me how  the inequality (2) was obtained. Any hlep would be appreciated.

If $\lambda = \frac 12,$ by using the inequality they provided :
$$\dfrac{\sqrt{n}}{a_n-a_{n-1}}>\frac 23\iff \sqrt{n} >\frac 23\left(a_n-a_{n-1}\right ).$$
Sum this up for and then you will get one side by telescoping. You can do a similar thing for $\lambda=\frac 34$ as well.
