Đặt A=\(\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\left(\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}\right)\)
Gọi \(\dfrac{a-b}{c}=x\); \(\dfrac{b-c}{a}=y\); \(\dfrac{c-a}{b}=z\) => \(\dfrac{c}{a-b}=\dfrac{1}{x};\dfrac{a}{b-c}=\dfrac{1}{y};\dfrac{b}{c-a}=\dfrac{1}{z}\)
=> A=(x+y+z)\(\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(=1+\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{y}{x}+1+\dfrac{y}{z}+\dfrac{z}{x}+\dfrac{z}{y}+1\)
= \(\dfrac{x+z}{y}+\dfrac{x+y}{z}+\dfrac{z+y}{x}+3\)
Lại có: \(\dfrac{z+y}{x}=\dfrac{\dfrac{b-c}{a}+\dfrac{c-a}{b}}{\dfrac{a-b}{c}}\) = \(\dfrac{b^2-bc+ac-a^2}{ab}.\dfrac{c}{a-b}\)= \(\dfrac{\left(b-a\right)\left(a+b\right)-c\left(b-a\right)}{ab}.\dfrac{c}{a-b}\) =\(\dfrac{\left(b-a\right)\left(a+b-c\right)}{ab}.\dfrac{c}{a-b}\) = \(\dfrac{-\left(a-b\right)\left(a+b-c\right)}{ab}.\dfrac{c}{a-b}\)= \(\dfrac{\left(-a-b+c\right).c}{ab}\) (1)
Lại có: a+b+c=0 <=> c=-a-b
Thay vào (1) ta được:\(\dfrac{z+y}{x}\)= \(\dfrac{2c^2}{ab}\)
Tương tự ta chứng minh được: \(\dfrac{x+z}{y}=\dfrac{2a^2}{bc}\) ; \(\dfrac{x+y}{z}=\dfrac{2b^2}{ac}\)
=> A=\(\dfrac{2a^2}{bc}+\dfrac{2b^2}{ac}+\dfrac{2c^2}{ab}\)+3 = \(\dfrac{2a^3}{abc}+\dfrac{2b^3}{abc}+\dfrac{2c^3}{abc}+3=\dfrac{2\left(a^3+b^3+c^3\right)}{abc}+3\)
ta chứng minh được \(a^3+b^3+c^3=3abc\) khi a+b+c=0
=> \(A=\dfrac{2.3abc}{abc}+3=6+3=9\)
Vậy A=9
<=> \(\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\left(\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}\right)\)=9
