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