Sử dụng IaI+IbI≥Ia+bI, Dấu = xảy ra khi ab≥0 ta được
A=Ix-2017I+Ix-2018I+Ix-2019I+Ix-2020I
=(I2020-xI+Ix-2017I)+(Ix-2018I+I2019-xI)
≥ I2020-x+x-2017I + Ix-2018+2019-xI =4
Vậy Min A=4 xảy ra khi $\left \{ {{(2020-x)(x-2017)\geq 0} \atop {(2019-x)(x-2018)\geq 0}} \right.$
$\Leftrightarrow \left \{ {{2020\geq x\geq 2017} \atop {2019\geq x\geq 2018}} \right.$
$\Leftrightarrow 2019\geq x\geq 2018$