$ \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 $x_i > 0$

Question

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

Answer 1

Let $x_0=0$.

Thus, by C-S $$\sum_{k=1}^n\frac{x_k}{1+\sum\limits_{i=1}^kx_i^2}\leq\sqrt{n\sum_{k=1}^n\frac{x_k^2}{\left(1+\sum\limits_{i=1}^kx_i^2\right)^2}}\leq$$ $$\leq\sqrt{n\sum_{k=1}^n\frac{x_k^2}{\left(1+\sum\limits_{i=0}^{k-1}x_i^2\right)\left(1+\sum\limits_{i=1}^kx_i^2\right)}}=\sqrt{n\left(\frac{\sum\limits_{k=1}^nx_k^2}{1+\sum\limits_{k=1}^nx_k^2}\right)}\leq\sqrt{n}.$$

Answer 2

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 $$\sum_{i=1}^{n} A_i < \sqrt{n}$$

Answer 3

Let $y_k=\sqrt{1+\sum_{h=1}^kx_h^2},\cos(\theta_k)=\frac{x_k}{y_k}$,

Since $y_k^2=y_{k-1}^2+x_k^2$, so $\sin(\theta_k)=\frac{y_{k-1}}{y_k}$

So $\frac{x_k}{1+\sum_{h=1}^kx_h^2}=\frac{\cos(\theta_k)}{y_k} = \frac{\sin(\theta_k)\cos(\theta_k)}{y_{k-1}}$

So LEFT = $\frac{\cos(\theta_1)}{y_1}+\frac{\cos(\theta_2)}{y_2}+...+\frac{\cos(\theta_n)}{y_n}$

$\frac{\cos(\theta_n)}{y_n} = \frac{\cos(\theta_n)\sin(\theta_n)}{y_{n-1}}\le \frac1{y_{n-1}}$

$\frac{\cos(\theta_{n-1})}{y_{n-1}}+\frac{\cos(\theta_n)}{y_n} \le \frac{\cos(\theta_{n-1})+1}{y_{n-1}}=\frac{\sin(\theta_{n-1})(1+\cos(\theta_{n-1}))}{y_{n-2}}\le\frac{\sin(\theta_{n-2})+\cos(\theta_{n-2})}{y_{n-2}}\le\frac{\sqrt{2}}{y_{n-2}}$

...

$\frac{\cos(\theta_1)}{y_1}+...+\frac{\cos(\theta_{n-1})}{y_{n-1}}+\frac{\cos(\theta_n)}{y_n} \le \frac{\cos(\theta_1)}{y_1}+\frac{\sqrt{n-1}}{y_1}=\frac{(\cos(\theta_1)+\sqrt{n-1})\sin(\theta_1)}{y_0}\le\frac{\cos(\theta_1)+\sqrt{n-1}\sin(\theta_1)}{y_0}\le\frac{\sqrt{n}}{y_0}=\sqrt{n}$