We have,
(a + b + c)(bc + ca + ab) > 9abc
= abc+ a2c + a2b + b2c+ abc + b2a + c2a + c2b + abc −9abc
= a (b2 + c2 ) + b(c2 + a2 ) + c(a2 + b2 ) − 6abc
= a (b2 + c2 − 2bc) + b(c2 + a2 − 2ca) + c(a2 + b2 − 2ab)
= a (b − c)2 + b(c − a)2 + c(a − b)2
Which is positive because each term of RHS is positive.
Thus,
(a + b + c)(bc + ca + ab) − 9abc > 0
(a + b + c)(bc + ca + ab) > 9abc.
