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Algebra Difficulty 4.9 AIME Prove it Saudi Arabia

For every real numbers $x_1, \dots, x_n$ , prove that
$\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2+x_2^2} + \dots + \frac{x_n}{1+x_1^2+\dots+x_n^2} < \sqrt{n}.$

Solution

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.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.