Verified answer

đặt:

A = b+c-a

B = a+c-b

C = b+a-c

=> a = (B+C)\2, b = (A+C)\2,c= (A+B)\2

== ====

ta có: (A\B + B\A) + (A\C + C\A) + (B\C+C\B) ≥ 6

=> (C+B)\A + (A+B)\C + (A+C)\B ≥ 6

=> 2a\(b+c-a) + 2b\(c+a-b) + 2c\(b+a-c) ≥ 6

=> đpcm