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Find k with $$\frac{1}{(3a-b)^2}+\frac{1}{(3b-c)^2}+\frac{1}{(3c-a)^2} \ge \frac{k}{a^2+b^2+c^2}$$

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    $k=-1$ will do for values where the left hand side and right hand side are both defined - or did you mean something different.2012-11-11
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    It seems likely that he's asking for the best possible $k$.2012-11-11
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    @Mark Bennet find all k, not just one2012-11-11

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