8

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 ?

Michael Hardy
  • 1
jannicehelm
  • 81

2 Answers2

12

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,

$\displaystyle \frac{x_1^2}{(1+x_1^2)^2}+\frac{x_2^2}{(1+x_1^2+x_2^2)^2}+\cdots+\frac{x_n^2}{(1+x_1^2+\cdots+x_n^2)^2} \le 1 - \frac{1}{1+x_1^2+\cdots+x_n^2} < 1$.

r9m
  • 18,208
  • I'd award you the bounty right now, but SE's policy doesn't allow me to do it for the next 22 hours. You're good ! (+1) – Victor Nov 05 '14 at 19:59
  • Ugh, I know this is super simple, but how did the $\sqrt{n}$ in the first step get converted to $n$? (Can you post a link that describes that first step in more detail?) – alfreema Nov 05 '14 at 21:56
  • @alfreema see that the LHS of the cauchy-schwarz has a square power .. you could also write that as $$\displaystyle \left(\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\cdots +\frac{x_n}{1+x_1^2+\cdots+x_n^2}\right) \le \sqrt{n}\left(\frac{x_1^2}{(1+x_1^2)^2}+\frac{x_2^2}{(1+x_1^2+x_2^2)^2}+\cdots+ \frac{x_n^2}{(1+x_1^2+ \cdots +x_n^2)^2}\right)^{1/2}$$ if you want to ! :) – r9m Nov 05 '14 at 22:03
-2

I think the proof can be divided to two parts: when all elements are equal, and then when they are not equal, try to prove that the sum would be less than sum of all equal elements (either max or min of the sequence).

First step: any x can be replaced by 1/x to increase the sum, if x > 1.

Second step: For x < y <= 1 and C >= 1, we have x / (C + x^2) < y / (C + y^2)

Third step: we replace all items with x(k) = max of sequence and have a more convenient formula to prove.

Fourth step: when all elements are equal, sum is less than 1/2 + 1/3 + 1/4 ... Which is the exact case when all elements are equal to 1, and this is a known sequence, as I remember

Timothy Ha
  • 101