Theo giải thiết thì ta có $a+b+c=2001 = n (n =2001)$
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c} =\frac{1}{n-c}+\frac{1}{n-a}+\frac{1}{n-b}= \frac{1}{10}$
đặt $n-c = x; n-b = y; n-a =z \to a = n-z; b = n-y; c = n-x$
hay
$\frac{1}{x}+\frac{1}{z}+\frac{1}{y} = \frac{1}{10}$
Từ đó $S = \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a}{n-a}+\frac{b}{n-b}+\frac{c}{n-c} = \frac{n-z}{z}+\frac{n-y}{y}+\frac{n-x}{x} = n(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})-3 = n\frac{1}{10}-3 = \frac{2001}{10}-3 =\frac{1971}{10}$