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

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!

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 $\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}+\cdots+\frac{1}{\sqrt{x_n}} \right )$

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 $\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}+\cdots+\frac{1}{\sqrt{x_n}} \right )$

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