Note that $(n+1)x_{n+1}-nx_n=x_n^2$, so summing gives $(n+1)x_{n+1}=2+x_1^2+...+x_n^2>\sum x_i^2 \geq \frac {(\sum x_i)^2}{n} $, done.

Here's a less smart solution: Let $S_n=x_1+\ldots+x_n$ for all $n$. We need to prove that $S_n < \sqrt{n(n+1)x_{n+1}}$. We procceed inductively. For $n=1$, the desired inequality obviously holds. Now, note that $S_{n+1}=S_n+x_{n+1}<\sqrt{n(n+1)x_{n+1}}+x_{n+1},$ and so it suffices to prove that $\sqrt{n(n+1)x_{n+1}}+x_{n+1}<\sqrt{(n+1)(n+2)x_{n+2}}$ However, $(n+2)x_{n+2}=x_{n+1}(x_{n+1}+n+1)$, and so we are left with proving $\sqrt{n(n+1)x_{n+1}}+x_{n+1}<\sqrt{(n+1)x_{n+1}(x_{n+1}+n+1)}$ Now, this dies to some easy algebraic manipulations, as it is equivalent to $2\sqrt{nx_{n+1} \cdot (n+1)}<nx_{n+1}+(2n+1),$ which is immediate from AM-GM and $2n+1>n+1$.