Điều kiện là ko có 2 số nào bằng nhau
Đặt \(\left(\frac{a}{b-c};\frac{b}{c-a};\frac{c}{a-b}\right)=\left(x;y;z\right)\)
\(\Rightarrow xy+yz+zx=\frac{ab}{\left(b-c\right)\left(c-a\right)}+\frac{bc}{\left(c-a\right)\left(a-b\right)}+\frac{ac}{\left(b-c\right)\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\)
\(\Rightarrow x^2+y^2+z^2=\left(x+y+z\right)^2-2\left(xy+yz+zx\right)=\left(x+y+z\right)^2+2\ge2\)
Dấu "=" xảy ra khi \(x+y+z=0\)
