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郝酒 发表于 2021-7-26 12:03
转一个微博上的解法：
1. Show that the inequality
\[
\sum_{i=1}^n \sum_{j=1}^n \sqrt{\left|x_i-x_j\right|} \leq \sum_{i=1}^n \sum_{j=1}^n \sqrt{\left|x_i+x_j\right|}
\]
holds for all real numbers $x_1, \ldots x_n$.
Solution:
For $a \inR^{+}$,
\[
\int \frac{1-\cos a x}{x \sqrt{x}} d x=4 \sqrt{a} S(\sqrt{a x})+\frac{2(\cos a x-1)}{\sqrt{x}}+C
\]
Let
\[
S(x)=\int_0^x \sin t^2 d t
\]
then
\[
S(0)=0, \quad S(+\infty)=\sqrt{\frac{\pi}{8}}
\]
Now
\[
\begin{gathered}
\sqrt{|a+b|}-\sqrt{|a-b|} \\
=\frac{1}{\sqrt{2 \pi}} \int_0^{+\infty} \frac{\cos (a-b) x-\cos (a+b) x}{x \sqrt{x}} d x \\
=\frac{1}{\sqrt{2 \pi}} \int_0^{+\infty} \frac{2 \sin a x \sin b x}{x \sqrt{x}} d x
\end{gathered}
\]
Rearranging,
\[
\begin{aligned}
& \sum_{i=1}^n \sum_{j=1}^n \sqrt{\left|x_i+x_j\right|}-\sqrt{\left|x_i-x_j\right|} \\
= & \sum_{i=1}^n \sum_{j=1}^n \frac{1}{\sqrt{2 \pi}} \int_0^{+\infty} \frac{2 \sin x_i t \sin x_j t}{t \sqrt{t}} d t \\
= & \frac{1}{\sqrt{2 \pi}} \int_0^{+\infty} \sum_{i=1}^n \sum_{j=1}^n \frac{2 \sin x_i t \sin x_j t}{t \sqrt{t}} d t \\
= & \frac{1}{\sqrt{2 \pi}} \int_0^{+\infty} \frac{2\left(\sum_{i=1}^n \sin x_i t\right)^2 d t}{t \sqrt{t}} \geq 0
\end{aligned}
\]
The equality holds when $\sum_{i=1}^n \sin(x_it)$ is identical $0$, which easily implies $\{ x_1,\ldots,x_n \}=\{ -x_1,\ldots,-x_n \}$.
这个解法的视频讲解： IMO 2021 Problem 2 - Hardest Inequality of IMO solved with An Amazing INTEGRAL
这个解法来源于AoPS：#58（一模一样）
#68 |
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hbghlyj
发表于 2024-4-24 18:11
