Xét hai trường hợp:
+ a + b + c \(\ne\) 0: Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a+b+c}{b+c+c+a+a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}a=2\left(b+c\right)\\b=2\left(c+a\right)\\c=2\left(a+\text{}b\right)\end{matrix}\right.\)
\(\Rightarrow a+b+c=2\left(b+c+c+a+a+b\right)\)
\(\Rightarrow a+b+c=4\left(a+b+c\right)\), vô lí vì a + b + c \(\ne\) 0.
+ a + b + c = 0: Khi đó \(\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{c}{a+b}=\dfrac{b+c}{-\left(b+c\right)}+\dfrac{a+c}{-\left(a+c\right)}+\dfrac{-\left(a+b\right)}{a+b}=-1+\left(-1\right)+\left(-1\right)=-3\)
