\(\Rightarrow\left(a+b+c\right).\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2010.\frac{1}{3}\)
Mà \(\left(a+b+c\right).\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=\)\(\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\)
\(=1+\frac{c}{a+b}+\frac{a}{b+c}+1+\frac{b}{c+a}+1=3+S\)
=> \(S=\frac{2010}{3}-3=\frac{2001}{3}\)