Ta có: (a-1)(b-1)(c-1) > 0 (*)
<=> (ab-a-b+1)(c-1) > 0
<=> abc - ab -ac +a -bc +b +c -1 >0
<=> 1 -ab - ac -bc +a+b+c -1 > 0
<=> -ab-ac-bc +a+b+c >0
<=> a+b+c > ab+ac+bc
<=> a+b+c > \(\dfrac{abc}{c}+\dfrac{abc}{b}+\dfrac{abc}{a}\)
<=> a+b+c > \(\dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{c}\) (1)
(1) đúng => (*) đúng
Vậy (a-1)(b-1)(c-1) > 0
Nhớ tick nhé ,chúc bạn học tốt