Find k with $$\frac{1}{(3a-b)^2}+\frac{1}{(3b-c)^2}+\frac{1}{(3c-a)^2} \ge \frac{k}{a^2+b^2+c^2}$$
Find k $\frac{1}{(3a-b)^2}+\frac{1}{(3b-c)^2}+\frac{1}{(3c-a)^2} \ge \frac{k}{a^2+b^2+c^2}$
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inequality
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1$k=-1$ 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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0It seems likely that he's asking for the best possible $k$. – 2012-11-11
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0@Mark Bennet find all k, not just one – 2012-11-11