Question regarding an inequality How to prove that $$ \frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\cdots+\frac{x_n}{1+x_1^2+\cdots+x_n^2}<\sqrt{n} $$ knowing that $(x_n)$ is a positive sequence ? I looked up all kinds of inequalities such AM-GM, Chebyshev, Cauchy-Schwarz, but I couldn't manage to obtain anything useful..
Can anyone help ? 

First use Cauchy-Schwarz Inequality:
$\displaystyle \left(\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\cdots+\frac{x_n}{1+x_1^2+\cdots+x_n^2}\right)^2 \le n\left(\frac{x_1^2}{(1+x_1^2)^2}+\frac{x_2^2}{(1+x_1^2+x_2^2)^2}+\cdots+\frac{x_n^2}{(1+x_1^2+\cdots+x_n^2)^2}\right)$
So, it suffices to show that: $\displaystyle \frac{x_1^2}{(1+x_1^2)^2}+\frac{x_2^2}{(1+x_1^2+x_2^2)^2}+\cdots+\frac{x_n^2}{(1+x_1^2+\cdots+x_n^2)^2} < 1$
Since, $\displaystyle \frac{x_k^2}{(1+x_1^2+\cdots+x_k^2)^2} \le \frac{x_k^2}{(1+x_1^2+\cdots+x_k^2)(1+x_1^2+\cdots+x_{k-1}^2)} = \frac{1}{(1+x_1^2+\cdots+x_{k-1}^2)} - \frac{1}{(1+x_1^2+\cdots+x_{k}^2)}$
and, $\displaystyle \frac{x_1^2}{(1+x_1^2)^2} \le 1- \frac{1}{(1+x_1^2)}$
Adding the above inequality, the upper bounds telescope,
$\displaystyle \frac{x_1^2}{(1+x_1^2)^2}+\frac{x_2^2}{(1+x_1^2+x_2^2)^2}+\cdots+\frac{x_n^2}{(1+x_1^2+\cdots+x_n^2)^2} \le 1 - \frac{1}{1+x_1^2+\cdots+x_n^2} < 1$.
