问题72 一道越南IMO2022试题

已知$a+b+c=2022$，$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2022}$，求$S=\dfrac{1}{a^{2023}}+\dfrac{1}{b^{2023}}+\dfrac{1}{c^{2023}}$

解析

$\because a+b+c=2022$

$\therefore \dfrac{1}{a+b+c}=\dfrac{1}{2022}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$

$\therefore\dfrac{1}{a+b+c}-\dfrac{1}{a}=\dfrac{1}{b}+\dfrac{1}{c}$

$\therefore \dfrac{-b-c}{a(a+b+c)}=\dfrac{b+c}{bc}$

$\therefore \dfrac{b+c}{bc}+\dfrac{b+c}{a(a+b+c)}=0$

$\therefore (b+c)\left[\dfrac{1}{bc}+\dfrac{1}{a(a+b+c)}\right]=0$

$\therefore (b+c)\times \dfrac{a^2+ab+ac+bc}{abc(a+b+c)}=0$

$\therefore \dfrac{(b+c)(a+c)(a+b)}{abc(a+b+c)}=0$

$\because abc(a+b+c) \neq 0$

$\therefore (a+b)(b+c)(c+a)=0$

不妨设$a+b=0$，则$c=2022$，且$a=-b$

$\therefore S=\dfrac{1}{(-b)^{2023}}+\dfrac{1}{b^{2023}}+\dfrac{1}{2022^{2023}}=\dfrac{1}{2022^{2023}}$