0
$\begingroup$

Let $0 \lt x_1 \le x_2 \le x_3 \le ... \le x_{n - 1} \le x_n$ and $n \geqslant 2$.

Also , $$\sum_\text{cyc} \frac{1}{x_1^2 + 2020^2} = \frac{1}{2020^2}$$

Prove that $$\frac{\sum_\text{cyc} x_1}{2020^2} \geq (n - 1)(\sum_{cyc} \frac{1}{x_1})$$

This is problem #5 (Afternoon) Thailand POSN Camp 2.

Can anyone give me any hints (or solutions) please. Thank you!

$\endgroup$
1
$\begingroup$

I have always noticed how they put the year in the question. I see that it is worthless. Write $2021$ in place of $2020$ and it won't change a thing :p \begin{align*} \sum_{cyc} \frac{1}{x_1^2 + 2020^2} &= \frac{1}{2020^2}\\ \sum_{cyc} \frac{1}{\frac{x_1^2}{2020^2}+1} &= 1\\ \sum_{cyc} \frac{1}{x^\prime_1+1} &= 1\\ \end{align*} where $$x^\prime_i=\frac{x_i^2}{2020^2}\Rightarrow \frac{x_i}{2020}=+\sqrt{x^\prime_i}\quad(\because x_i>0)$$.

So, we need to prove \begin{align*} \frac{\sum_\text{cyc} x_1}{2020} &\geq (n - 1)(\sum_{cyc} \frac{2020}{x_1})\\ \text{or}\quad\sum_\text{cyc} \frac{x_1}{2020} &\geq (n - 1)(\sum_{cyc} \frac{2020}{x_1})\\ \text{or}\quad\sum_\text{cyc} \sqrt{x^\prime_1} &\geq (n - 1)(\sum_{cyc} \frac{1}{\sqrt{x^\prime_1}})\\ \end{align*}
Now, Prove that $\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}+\cdots+\frac{1}{\sqrt{x_n}} \right )$

$\endgroup$
1
$\begingroup$

Your inequality is true for any positive variables such that:$$\sum_{cyc} \frac{1}{x_1^2 + 2020^2} = \frac{1}{2020^2}.$$

Indeed, let $a_i=\frac{x_i^2}{2020}$ and see here: Prove that $\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}+\cdots+\frac{1}{\sqrt{x_n}} \right )$

$\endgroup$
4
  • $\begingroup$ Can you give more specific detail please. $\endgroup$ – dark.nes_s Jun 22 '20 at 14:00
  • $\begingroup$ Do we have to use Chebyshev? Sorry for my stupidity. $\endgroup$ – dark.nes_s Jun 22 '20 at 14:00
  • $\begingroup$ @Ivar the Boneless Yes we can. See my post in the linked topic. $\endgroup$ – Michael Rozenberg Jun 22 '20 at 15:52
  • $\begingroup$ Get it! Thank you very much. $\endgroup$ – dark.nes_s Jun 23 '20 at 1:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.