Cho a+b+c=2014 và 1/(a+b)+1/(b+c)+1/(c+a)=1/2014 Tình S= c/(a+b)+a/(b+c)+b/(c+a)

$\text{S=$\frac{a}{b+c}$ + $\frac{b}{c+a}$ + $\frac{c}{a+b}$}$

$\text{⇔ S+3=$\frac{a}{b+c}$+1+$\frac{b}{c+a}$+1+$\frac{c}{a+b}$+1}$

$\text{⇔ S+3=$\frac{a}{b+c}$+$\frac{b+c}{b+c}$+$\frac{b}{c+a}$+$\frac{c+a}{c+a}$+$\frac{c}{a+b}$+$\frac{a+b}{a+b}$}$

$\text{⇔ S+3=$\frac{a+b+c}{b+c}$+$\frac{b+c+a}{c+a}$+$\frac{c+a+b}{a+b}$}$

$\text{⇔ S+3=(a+b+c)($\frac{1}{b+c}$ + $\frac{1}{c+a}$ + $\frac{1}{a+b}$)}$

$\text{Thay a+b+c=2014 và ($\frac{1}{b+c}$ + $\frac{1}{c+a}$ + $\frac{1}{a+b}$)=$\frac{1}{2014}$ ta được}$

$\text{S+3=2014.$\frac{1}{2014}$ }$

$\text{⇔S+3=1}$

$\text{⇔S=-2}$