Vietnamese Mathematical Competitions

We have $x_{n+1} < x_{n+1} + \frac{1}{4n^4} = x_n + 2\frac{\sqrt{x_n}}{2n^2} + \frac{1}{4n^4} = \left(\sqrt{x_n} + \frac{1}{2n^2}\right)^2$.
Therefore $\sqrt{x_{n+1}} < \sqrt{x_n} + \frac{1}{2n^2}$, $n = 1, 2, 3, \dots$ thus
$$\begin{aligned} \sqrt{x_{n+1}} &< \sqrt{a} + \sum_{i=1}^{n} \frac{1}{2i^2} < \sqrt{a} + \frac{1}{2} + \sum_{i=2}^{n} \frac{1}{2i(i-1)} < \sqrt{a} + \frac{1}{2} + \frac{1}{2} \sum_{i=2}^{n} \left( \frac{1}{i-1} - \frac{1}{i} \right) = \\ &= \sqrt{a} + \frac{1}{2} + \frac{1}{2} \left( 1 - \frac{1}{n} \right) < \sqrt{a} + 1 \end{aligned}$$
So $x_{n+1} < (\sqrt{a} + 1)^2$ for all $n$. The given sequence is bounded above, strictly increasing and so has a finite limit. (Q.E.D).