\(S=\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{a+b}\)
\(3+S=1+\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}\)
\(3+S=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(3+S=\left(a+b+c\right).\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(3+S=\frac{2001.1}{10}=\frac{2001}{10}\Rightarrow S=\frac{2001}{10}-3\)