cho a.b.c =1 . chứng minh rằng 1/ab+a +1 . 1/bc+b+1 + 1/ca+c+1 > 2019/2020

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1. 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 $\dfrac{1}{ab+a+1} + \dfrac{1}{bc+b+1}+\dfrac{1}{ca+c+1} > \dfrac{2019}{2020}$

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