\(a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
⇔ \(a+b+c=\dfrac{abc}{a}+\dfrac{abc}{b}+\dfrac{abc}{c}\)
(do abc = 1)
⇔ a + b +c = bc + ac + ab
Ta có:
A = (a - 1)(b - 1)(c - 1)
= (ab - a - b + 1)(c - 1)
= abc - ab - ac - bc + a + b + c - 1
= 1 - ab - ac - bc + bc + ac + ab - 1 (do abc = 1; a + b + c = bc + ac + ab)
= 0
Vậy A = 0
Ta có 1/a + 1/b + 1/c = (bc + ac + ac)/abc = ab + bc + ca
=> a + b + c = ab + bc + ca
<=> a + b + c - ab - bc - ca = 0
<=> a + b + c - ab - bc - ac + abc - 1 = 0
<=> (a - ab) + (b - 1) + (c - bc) + (abc - ac) = 0
<=> -a(b - 1) + (b - 1) - c(b - 1) + ac(b - 1) = 0
<=> (b - 1)(-a + 1 -c + ac) = 0
<=> (b - 1)[ (-a + 1) + (ac - c) ] = 0
<=> (b - 1)[ -(a - 1) + c(a - 1) ] = 0
<=> (a - 1)(b - 1)(c - 1) = 0
Vậy....