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

Question

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

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 $$2460640\,{z}^{3}-2787824\,{z}^{2}+905866\,z-210681$$