Có a+b+c = 20170
=> a= 20170- (b+c)
b= 20170 - (a+c)
c= 20170 - (a+b)
S = $\frac{a}{b+c}$ + $\frac{b}{a+c}$ +$\frac{c}{a+b}$
= $\frac{20170- (b+c)}{b+c}$ + $\frac{ 20170 - (a+c)}{a+c}$+$\frac{20170 - (a+b)}{a+b}$
= $\frac{20170}{b+c}$- 1+ $\frac{20170}{a+c}$- 1+ $\frac{20170}{a+b}$- 1
= 20170( $\frac{1}{b+c}$ + $\frac{1}{a+c}$ +$\frac{1}{a+b}$)-3
= 20170. $\frac{1}{10}$- 3
= 2017 -3= 2014