dễ thấy a+b; b+c; c+a khác 0
suy ra: [tex](a-b).(b-c).(c-a)=(a+b).(b+c).(c+a)\\\\ <=> \frac{a-b}{a+b}.\frac{b-c}{b+c}.\frac{c-a}{c+a}=1[/tex]
đặt: [tex]\frac{a-b}{a+b}=x\\\\ \frac{b-c}{b+c}=y\\\\ \frac{c-a}{c+a}=z\\\\ => xyz=1\\\\ +, x+1=\frac{2a}{a+b}; y+1=\frac{2b}{b+c};z+1=\frac{2c}{c+a}\\\\ +, 1-x=\frac{2b}{a+b};1-y=\frac{2c}{b+c}; 1-z=\frac{2a}{c+a}\\\\ => (x+1).(y+1).(z+1)=(1-x).(1-y).(1-z)\\\\ <=> (x+1).(y+z+1+yz)=(1-x).(1-y-z+yz)\\\\ <=> xy+xz+x+xyz+y+z+1+yz=1-y-z+yz-x+xy+xy-xyz\\\\ <=> 2x+2y+2z=-2xyz=-2\\\\ <=> x+y+z=-2\\\\ <=> x+1 + y+1 + z+1 =1\\\\ <=> \frac{2a}{a+b}+\frac{2b}{b+c}+\frac{2c}{c+a}=1\\\\ => A=\frac{1}{2}[/tex]