Đáp án:
$\dfrac{2018}{2019} + \dfrac{2019}{2020} + \dfrac{2020}{2021} +\dfrac{2021}{2018} > 4$
Giải thích các bước giải:
Ta có:
$\dfrac{2018}{2019} + \dfrac{2019}{2020} + \dfrac{2020}{2021} +\dfrac{2021}{2018} - 4$
$= \left(\dfrac{2018}{2019} -1\right) + \left(\dfrac{2019}{2020} -1\right)+\left(\dfrac{2020}{2021} -1\right)+\left(\dfrac{2021}{2018} -1\right)$
$= -\dfrac{1}{2019} -\dfrac{1}{2020} -\dfrac{1}{2021} +\dfrac{3}{2018}$
$= \left(\dfrac{1}{2018} -\dfrac{1}{2019}\right)+ \left(\dfrac{1}{2018} -\dfrac{1}{2020}\right)+\left(\dfrac{1}{2018} -\dfrac{1}{2021}\right) > 0$
Do đó:
$\dfrac{2018}{2019} + \dfrac{2019}{2020} + \dfrac{2020}{2021} +\dfrac{2021}{2018} > 4$
