Giải thích các bước giải:
Ta có :
$\dfrac{a^{2020}}{b^{2019}}+b+b+..+b\ge 2020\sqrt[2020]{\dfrac{a^{2020}}{b^{2019}}.b.b..b}$(2019 b)
$\to \dfrac{a^{2020}}{b^{2019}}+2019b\ge 2020a$
Chứng minh tương tự
$\to \dfrac{b^{2020}}{c^{2019}}+2019c\ge 2020b$
$\to \dfrac{c^{2020}}{a^{2019}}+2019a\ge 2020c$
$\to \dfrac{a^{2020}}{b^{2019}}+\dfrac{b^{2020}}{c^{2019}}+\dfrac{c^{2020}}{a^{2019}}+2019(a+b+c)\ge 2020(a+b+c) $
$\to \dfrac{a^{2020}}{b^{2019}}+\dfrac{b^{2020}}{c^{2019}}+\dfrac{c^{2020}}{a^{2019}}\ge a+b+c $
