Find the limit of $x_n^3/n^2$ if $x_{n+1}=x_{n}+1/\sqrt{x_n}$

The associated differential equation is $\xi'(t)=\frac1{\sqrt{\xi(t)}}$, whose solutions are $\xi(t)^{3/2}=\frac32t+C$. This suggests to look at the sequence $(z_n)$ defined by $z_n=x_n^{3/2}$. Thus, $$ z_{n+1}=z_n(1+z_n^{-1})^{3/2}. $$ First, $(1+u)^{3/2}\geqslant1+\frac32u$ for every $u\geqslant0$ hence $z_{n+1}\geqslant z_n+\frac32$, thus $z_n\geqslant\frac32n+z_0$ and in particular $z_n\to+\infty$.

On the other hand, $(1+u)^{3/2}\leqslant1+\frac32u+\frac38u^2$ for every $u\geqslant0$ hence $z_{n+1}\leqslant z_n+\frac32+\frac38z_n^{-1}$. This shows that $z_n\leqslant\frac32n+z_0+\frac38t_n$ with $t_n=\sum\limits_{k=0}^{n-1}z_k^{-1}$. Since $z_n\to+\infty$, $t_n=o(n)$ and $z_n\leqslant\frac32n+o(n)$. Finally, $\frac{z_n}n\to\frac32$ hence $$\lim\limits_{n\to\infty}\frac{x_n^3}{n^2}=\frac94. $$