$x_n \ge \frac{(n+1)(n+2)}{6}$

This is true for $n=1$. Now suppose $n \ge 2$ and suppose that, inductively:
$$x_{n-1} \ge \frac{n(n+1)}{6}$$
Now as $n \ge 2$, we have:
$$\frac{n+1}{n} = 1 + \frac{1}{n} \le \frac{3}{2}$$
Multiplying each side by $\frac{n(n+1)}{9}$:
$$\frac{(n+1)^2}{9} \le \frac{n(n+1)}{6}$$
As both sides are positive, we can take the square root of each side, which gives:
$$\frac{n+1}{3} \le \sqrt{\frac{n(n+1)}{6}}$$
This allows us to complete the inductive step:
$$x_n = x_{n-1} + \sqrt{x_{n-1}} \\ \ge \frac{n(n+1)}{6} + \sqrt{\frac{n(n+1)}{6}} \\ \ge \frac{n(n+1)}{6} + \frac{n+1}{3} = \frac{(n+1)(n+2)}{6}.$$

This is the inductive step we had to prove. Now we can apply this to the sum, which is:
$$\sum_{n=1}^{2018} \frac{1}{x_n} \le \sum_{n=1}^{2018} \frac{6}{(n+1)(n+2)} = \sum_{n=1}^{2018} \left( \frac{6}{n+1} - \frac{6}{n+2} \right ) = 3 - \frac{6}{2020} < 3.$$