1.

If a b, c are positive and not all equal, then prove that: (a + b + c)(bc + ca + ab) > 9abc

Answer»

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.



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