(1/a-1)(1/b-1)(1/c-1)=[(1-a)/a][(1-b)/b][(1-c)/c]=[(b+c)/a][(c+a)/b][(a+b)/c],由均值不等式:(a+b)(b+c)(c+a)>=[2√(bc)]*[2√(ac)]*[2√(ab)]=8abc所以(1/a-1)(1/b-1)(1/c-1)>=(8abc)/(abc)=8得证....