Thay $1=abc$ vào biểu thức ta có :
$\dfrac{1}{ab+a+1} + \dfrac{1}{bc+b+1}+\dfrac{1}{ca+c+1}$
$ = \dfrac{1}{ab+a+abc}+\dfrac{1}{bc+b+1}+\dfrac{1}{ca+c+abc}$
$ = \dfrac{abc}{a.(b+1+bc)}+\dfrac{1}{bc+b+1} + \dfrac{1}{c.(a+1+ab)}$
$ = \dfrac{bc}{b+1+bc}+\dfrac{1}{bc+b+1}+\dfrac{abc}{c.(a+abc+ab)}$
$ = \dfrac{bc+1}{b+1+bc}+\dfrac{abc}{ca.(1+b+bc)}$
$= \dfrac{bc+1}{b+1+bc}+\dfrac{b}{1+b+bc}$
$ = \dfrac{bc+1+b}{b+1+bc} = 1 > \dfrac{2019}{2020}$
Vậy $\dfrac{1}{ab+a+1} + \dfrac{1}{bc+b+1}+\dfrac{1}{ca+c+1} > \dfrac{2019}{2020}$
