$\begin{array}{l} \dfrac{a}{b} = \dfrac{{a\left( {b + 2021} \right)}}{{b\left( {b + 2021} \right)}} = \dfrac{{ab + 2021a}}{{b\left( {b + 2021} \right)}}\\ \dfrac{{a + 2021}}{{b + 2021}} = \dfrac{{b\left( {a + 2021} \right)}}{{b\left( {b + 2021} \right)}} = \dfrac{{ab + 2021b}}{{b\left( {b + 2021} \right)}} \end{array}$
Vì $b>0$ nên mẫu số của hai phân số dương nên ta chỉ cần so sánh tử số.
Nếu $a<b\Rightarrow ab+2021a<ab+2021b\Rightarrow \dfrac{a}{b}<\dfrac{a+2021}{b+2021}$
Nếu $a=b\Rightarrow ab+2021a=ab+2021b\Rightarrow \dfrac{a}{b}=\dfrac{a+2021}{b+2021}$
Nếu $a>b\Rightarrow ab+2021a>ab+2021b\Rightarrow \dfrac{a}{b}>\dfrac{a+2021}{b+2021}$
