• Nếu $a+b+c=0 \Longrightarrow \left\{\begin{matrix} a+b=-c\\b+c=-a\\ c+a=-b \end{matrix}\right.$
Nên
$T= \dfrac{a}{b+c}+ \dfrac{b}{a+c}+\dfrac{c}{a+b}=\dfrac{a}{-a}+\dfrac{b}{-b}+\dfrac{c}{-c}=-1-1-1=-3$
• Nếu $a+b+c \ne 0$
Áp dụng tính chất DTSBN, ta có:
$\dfrac{2014a+b+c}{a}=\dfrac{a+2014b+c}{b}=\dfrac{ a+b+2014c}{c}=\dfrac{2014a+b+c+2014b+c+a+2014c+b+a}{a+b+c}=\dfrac{2016(a+b+c)}{a+b+c}=2016$
$\Longrightarrow \dfrac{2014a+b+c}{a}=2016 \Longrightarrow 2014a+b+c=2016a \Longrightarrow b+c=2a$
Tương tự có: $a+c=2b; a+b=2c$
Thay vô T rồi tính