Ta có
\(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}\)
\(=\dfrac{abc}{ab+b+abc}+\dfrac{abc}{bc+c+abc}+\dfrac{1}{ca+a+1}\)
\(=\dfrac{abc}{b\left(ac+a+1\right)}+\dfrac{abc}{c\left(ab+b+1\right)}+\dfrac{1}{ac+a+1}\)
\(=\dfrac{ac}{ac+a+1}+\dfrac{ab}{ab+b+1}+\dfrac{1}{ac+a+1}\)
\(=\dfrac{ac+1}{ac+a+1}+\dfrac{ab}{ab+b+abc}\)
\(=\dfrac{ac+1}{ac+a+1}+\dfrac{ab}{b\left(ac+a+1\right)}=\dfrac{ac+a+1}{ac+a+1}=1\) (đpcm)
Ta có: \(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}\)
=\(\dfrac{1}{ab+b+1}+\dfrac{abc}{bc+c+abc}+\dfrac{b}{abc+ab+b}\)
=\(\dfrac{1}{ab+b+1}+\dfrac{abc}{c\left(ab+b+1\right)}+\dfrac{b}{ab+b+1}\)
=\(\dfrac{1}{ab+b+1}+\dfrac{ab}{ab+b+1}+\dfrac{b}{ab+b+1}\)
=\(\dfrac{ab+b+1}{ab+b+1}\)=1
Suy ra:
\(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}\)=1(abc=1)
(đpcm)
