Saudi Arabia Mathematical Competitions 2012

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