Ta có:
$\dfrac{1}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{1}{abc+bc+b}$
$=\dfrac{abc}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{1}{1+bc+b}$ (do $abc=1$)
$=\dfrac{abc}{a(bc+b+1)}+\dfrac{b}{bc+b+1}+\dfrac{1}{1+bc+b}$
$=\dfrac{bc}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{1}{1+bc+b}$
$=\dfrac{bc+b+1}{bc+b+1}=1$
(đpcm)