\(S=\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ca}\)
=\(\frac{c}{c\left(1+a+ab\right)}+\frac{ca}{ca\left(1+b+bc\right)}+\frac{1}{1+c+ca}\)
=\(\frac{c}{c+ca+abc}+\frac{ca}{ca+abc+abc.c}+\frac{1}{1+c+ca}\)
thay abc = 1 ta được
\(S=\frac{c}{c+ca+1}+\frac{ca}{ca+1+c}+\frac{1}{1+c+ca}\)(cùng mẫu c+ca+1)
\(=\frac{c+ac+1}{c+ac+1}=1\)