Question

Chờ a,b,c >0 có tích a.b.c =1 va a+b+c>\(\dfrac{1}{a}\)+\(\dfrac{1}{b}\)+\(\dfrac{1}{c}\)

Chứng minh rằng : (a -1).(b -1).(c -1)>0

Answer 1

Từ (a-1)(b-1)(c-1)>0 (*)

<=>(ab-b-a+1)(c-1)>0

<=> abc-ab-bc+b-ac+a+c-1>0

<=> a+b+c-ab-ac-bc>0

<=> a+b+c-\(\dfrac{abc}{c}-\dfrac{abc}{b}-\dfrac{abc}{a}\)>0

<=> a+b+c - \(\dfrac{1}{c}-\dfrac{1}{b}-\dfrac{1}{a}>0\)

<=> \(a+b+c>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ( 1)

(1) đúng => (*) đúng