Có: $x=\frac{a-b}{a+b}\Rightarrow \begin{cases}1+x=\frac{(a+b)+(a-b)}{a+b}=\frac{2a}{a+b} \\ 1-x=\frac{(a+b)-(a-b)}{a+b}=\frac{2b}{a+b} \end{cases}$Chứng minh tương tự: $1+y=\frac{2b}{b+c} ; 1-y=\frac{2c}{b+c}$
$1+z=\frac{2c}{c+a} ; 1-z=\frac{2a}{c+a}$
$\Rightarrow (1+x)(1+y)(1+z)=\frac{8abc}{(a+b)(b+c)(c+a)}$
$(1-x)(1-y)(1-z)=\frac{8abc}{(a+b)(b+c)(c+a)}$
$\Rightarrow (1+x)(1+y)(1+z)=(1-x)(1-y)(1-z)$
Đúng thì nhấp V nha.