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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. –  Mark Bennet Nov 11 '12 at 6:49
    
It seems likely that he's asking for the best possible $k$. –  user27126 Nov 11 '12 at 6:54
    
@Mark Bennet find all k, not just one –  Xeing Nov 11 '12 at 6:58
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1 Answer 1

In other words, you want to minimize $$f(a,b,c) = (a^2 + b^2 + c^2)\left(\frac{1}{(3a-b)^2} + \frac{1}{(3b-c)^2} + \frac{1}{(3c-a)^2}\right)$$ (presumably for $a,b,c$ real numbers). By homogeneity, we may restrict to the sphere $a^2 + b^2 + c^2 = 1$. Using a Lagrange multiplier, we look for critical points of $F(a,b,c,\lambda) = f(a,b,c) + \lambda (a^2 + b^2 + c^2-1)$. The solution is rather unpleasant. It seems the minimum is approximately $0.808645463814078870$.

EDIT: In fact this number turns out to be the real root of the polynomial $$2460640\,{z}^{3}-2787824\,{z}^{2}+905866\,z-210681$$

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why do we need Lagrange multiplier, there're no constraints? –  user31280 Nov 11 '12 at 8:58
    
Without the restriction, you get critical rays rather than a finite number of critical points. –  Robert Israel Nov 11 '12 at 20:10
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