\(\Leftrightarrow\left(1+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}\right)\left(1+\frac{c}{b}\right)=8\)
\(\Leftrightarrow2+\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}=8\)
\(\Leftrightarrow\frac{\left(a^2+b^2\right)c+\left(c^2+a^2\right)b+\left(b^2+c^2\right)a}{abc}=6\)
\(\Leftrightarrow a^2b+ab^2+b^2c+bc^2+a^2c+ac^2=6abc\)(1)
\(\Leftrightarrow\left(a^2b-2abc+bc^2\right)+\left(ab^2-2abc+ac^2\right)+\left(a^2c-2abc+b^2c\right)=0\)
\(\Leftrightarrow b\left(a-c\right)^2+a\left(b-c\right)^2+c\left(a-b\right)^2=0\)
Mà a,b,c khác 0 nên a=b=c
