\[\frac{a}{1+9bc+K(b-c)^2}+\frac{b}{1+9ca+K(c-a)^2}+\frac{c}{1+9ab+K(a-b)^2}\geq \frac{1}{2}\]

Find the maximum value of \( K \), such that for all non-negative numbers that satisfy \( a+b+c = 1 \), the above inequality is true.

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