\(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc+\left(a+b+c\right)-\left(ac+bc+ac\right)-1\)
\(\left\{{}\begin{matrix}abc=1\left(1\right)\\a+b+c>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(2\right)\end{matrix}\right.\) lấy (2) nhân (1) \(\Rightarrow a+b+c>ab+bc+ac\Leftrightarrow\left(a+b+c\right)-\left(ab+bc+ac\right)>0\) (3)
Thay (1) vào A \(\Leftrightarrow A=a+b+c-\left(ac+bc+ac\right)\)
Từ (3) => A>0 => dpcm
Phân tích :
(a-1)(b-1)(c-1) > 0 (*)
<=> (ab-a-b+1)(c-1)>0
<=> abc - ab-ac+a-bc+b +c-1>0
<=> a+b+c -ab-ac-bc >0
<=> \(a+b+c-\dfrac{abc}{c}-\dfrac{abc}{b}-\dfrac{abc}{a}>0\)
\(\Leftrightarrow a+b+c>\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a}\) (1)
(1) => (*) đúng (đpcm)
