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