<s>[INST] $ \frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2+x_2^2} +...+\frac{x_n}{1+x_1^2+x_2^2+...x_n^2} \le \sqrt{n}$ for $x_i > 0$ If $ x_1, x_2 , x_3........x_n $ are n positive reals prove that 
$$ \frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2+x_2^2} +...+\frac{x_n}{1+x_1^2+x_2^2+...x_n^2} \le \sqrt{n}$$
So this is an IMO 2001 proposed question.It was in the section of Cauchy Schwartz inequalities in my book. It seems that there is a Cauchy hidden because of the root n but then we need to find two series whose terms are those of the LHS and the product of sum of whose squares gives n . I'm at a loss in finding that
Thanks for any help
 [/INST] Let us call the individual terms of the inequality as $A_i$, then by AM-RMS inequality we have
$$\frac{\sum A_i}{n} \le \sqrt{\frac{\sum A_i^2}{n}},~\mbox{where} ~A_i=\frac{x_i}{1+x_1^2+x_2^2+x_3^2+...x_i^2}.~~~~(1)$$
So it would suffice to prove that $\sum A_i^2 \le 1.$
Note that for $i\ge 2$, $$ A_i^2=\frac{x_1^2}{(1+x_1^2+x_3^2+x_3^2+...x_i^2)^2}\le  \frac{x_i^2}{(1+x_1^2+x_3^2+x_3^2+...x_{i-1}^2)~ (1+x_1^2+x_3^2+x_3^2+...x_{i}^2)}.$$
$$=\frac{1}{1+x_1^2+x_2^2+x_3^2+...+x_{i-1}^2}-\frac{1}{1+x_1^2+x_2^2+x_3^2+...+x_{i}^2}.$$
We also have $$\frac{x_1^2}{1+x_1^2} \le 1- \frac{1}{1+x_1^2}.$$
Summing all these we get $$\sum_{i=1}^{n} A_i^2 \le 1-\frac{1}{1+x_1^2+x_2^2+x_3^2+...+x_{i}^2}<1.$$
Finally from (1) it follows that $$\sum_{i=1}^{n} A_i < \sqrt{n}$$
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