How prove $\frac{\sqrt{2}}{3}n^2<\sum_{k=1}^{n^2-1}\sqrt{1-\frac{\sqrt{k}}{n}}<\sqrt{2}n^2$ 
Show  that
  $$\dfrac{\sqrt{2}}{3}n^2<\sqrt{1-\dfrac{\sqrt{1}}{n}}+\sqrt{1-\dfrac{\sqrt{2}}{n}}+\sqrt{1-\dfrac{\sqrt{3}}{n}}+\cdots+\sqrt{1-\dfrac{\sqrt{n^2-1}}{n}}<\sqrt{2}n^2.$$

Maybe use 
$$(1+x)^{\frac{1}{2}}<1+\dfrac{x}{2},x>-1$$
so
$$\sqrt{1-\dfrac{\sqrt{k}}{n}}<1-\dfrac{1}{2}\cdot\dfrac{\sqrt{k}}{n}$$
then how to prove it?
I don't want see without Mathematical induction,maybe can use integral to prove it?
 A: Note that $f(x)=\sqrt{1-\sqrt{x}}~$ is strictly decreasing on $[0,1]$. Thus
for $0\leq k<m$ we have
$$
\int_{k/m}^{(k+1)/m}f(t)dt<\frac{1}{m}f\left(\frac{k}{m}\right)\tag{1}
$$
and for for $0< k\leq m$ we have
$$
\int_{(k-1)/m}^{k/m}f(t)dt>\frac{1}{m}f\left(\frac{k}{m}\right)\tag{2}
$$
So, if 
$$I_m=\frac{1}{m}\sum_{k=1}^{m-1}\sqrt{1-\sqrt{\frac{k}{m}}}$$ then,
from $(1)$ we get
$$
\int_{1/m}^1f(t)dt< I_m<\int_{0}^{1-1/m}f(t)dt<\int_{0}^{1}f(t)dt
$$
Now, it is not difficult to see that 
$$\left(-\frac{4}{15}(1-\sqrt{x})^{3/2}(2+3\sqrt{x})\right)'=f(x)$$
Thus
$$
\frac{4}{15}\left(1-\frac{1}{\sqrt{m}}\right)^{3/2}\left(2+ \frac{3}{\sqrt{m}}\right)<I_m<\frac{8}{15}
$$
In particular, for $m=n^2$ we have
$$
\frac{4}{15}\left(1-\frac{1}{n}\right)^{3/2}\left(2+ \frac{3}{n}\right)<I_{n^2}<\frac{8}{15}
$$
Now, the left hand side of the previous inequality is an increasing function of $n$. So, for $n\geq 4$ we have
$$
\frac{\sqrt{2}}{3}<\frac{11\sqrt{3}}{40}<I_{n^2}<\frac{8}{15}
$$
Now, a direct check shows that the proposed inequality is valid also for $n=3$, but it is not for $n=1,2$. So, we have proved that
So, for $n\geq 3$ we have
$$
n^2\frac{\sqrt{2}}{3} <\sum_{k=1}^{n^2-1}\sqrt{1- \frac{\sqrt{k}}{n}}<\frac{8}{15}n^2
$$
which is the desired conclusion, with a stronger upper bound.
