$\begin{array}{l} \dfrac{a}{b} = \dfrac{{a\left( {b + 2021} \right)}}{{b\left( {b + 2021} \right)}} = \dfrac{{ab + 2021a}}{{b\left( {b + 2021} \right)}}\\ \dfrac{{a + 2021}}{{b + 2021}} = \dfrac{{b\left( {a + 2021} \right)}}{{b\left( {b + 2021} \right)}} = \dfrac{{ab + 2021b}}{{b\left( {b + 2021} \right)}}\\ \left( {do\,b > 0 \Rightarrow b\left( {b + 2021} \right) > 0} \right)\\  + a < b \Rightarrow \dfrac{a}{b} = \dfrac{{ab + 2021a}}{{b\left( {b + 2021} \right)}} < \dfrac{{ab + 2021b}}{{b\left( {b + 2021} \right)}} = \dfrac{{a + 2021}}{{b + 2021}}\\  + b < a \Rightarrow \dfrac{a}{b} = \dfrac{{ab + 2021a}}{{b\left( {b + 2021} \right)}} > \dfrac{{ab + 2021b}}{{b\left( {b + 2021} \right)}} = \dfrac{{a + 2021}}{{b + 2021}}\\  + b = a \Rightarrow \dfrac{a}{b} = \dfrac{{ab + 2021a}}{{b\left( {b + 2021} \right)}} = \dfrac{{ab + 2021b}}{{b\left( {b + 2021} \right)}} = \dfrac{{a + 2021}}{{b + 2021}} \end{array}$

Ta có:

`+) a/b - 1 = a/b - b/b = (a - b)/b`

`+) (a + 2021)/(b + 2021) - 1 = (a + 2021)/(b + 2021) - (b + 2021)/(b + 2021)`

`= ((a + 2021) - (b + 2021))/(b + 2021) = (a + 2021 - b - 2021)/(b + 2021)`

`= (a - b)/(b + 2021)`

Vì `a > b`

`=> a - b > 0`

Vì `2021 > 0`

`=> b + 2021 > b`

`=> 1/(b + 2021) < 1/b`

`=> (a - b)/(b + 2021) < (a - b)/b` (Do `a - b > 0`)

`=> (a + 2021)/(b + 2021) - 1 < a/b - 1`

`=> (a + 2021)/(b + 2021) < a/b` hay `a/b > (a + 2021)/(b + 2021)`
Vậy `a/b > (a + 2021)/(b + 2021)` khi `a > b > 0`