Saudi Arabian Mathematical Competitions, 2012

Using Cauchy-Schwarz inequality we have
$$\begin{aligned} & \left( \frac{x_1}{1+x_1^2} + \dots + \frac{x_n}{1+x_1^2 + \dots + x_n^2} \right) \\ & \le n \left[ \frac{x_1^2}{(1+x_1^2)^2} + \dots + \frac{x_n^2}{(1+x_1^2 + \dots + x_n^2)^2} \right] \\ & \le n \left[ \frac{x_1^2}{1 \cdot (1+x_1^2)} + \frac{x_2^2}{(1+x_1^2)(1+x_1^2+x_2^2)} + \dots + \frac{x_n^2}{(1+x_1^2+\dots+x_{n-1}^2)(1+x_1^2+\dots+x_n^2)} \right] \\ & = n \left[ 1 - \frac{1}{1+x_1^2} + \frac{1}{1+x_1^2} - \frac{1}{1+x_1^2+x_2^2} + \dots + \frac{1}{1+x_1^2+\dots+x_{n-1}^2} - \frac{1}{1+x_1^2+\dots+x_n^2} \right] \\ & = n \left( 1 - \frac{1}{1+x_1^2+\dots+x_n^2} \right) < n, \end{aligned}$$
and the inequality follows.