a) Ta có: [tex]\left\{\begin{matrix} a-b=\frac{1}{c}-\frac{1}{b}=\frac{b-c}{bc}\\ b-c=\frac{1}{a}-\frac{1}{c}=\frac{c-a}{ca}\\ c-a=\frac{1}{b}-\frac{1}{a}=\frac{a-b}{ab} \end{matrix}\right.\Rightarrow (a-b)(b-c)(c-a)=\frac{(a-b)(b-c)(c-a)}{a^2b^2c^2}\Rightarrow a^2b^2c^2=1\Rightarrow P=\pm x[/tex]
b) Theo BĐT Cauchy ta có: [tex]9=(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9\Rightarrow x=y=z=3\Rightarrow T=162[/tex]