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$\sum \frac{a}{1+3bc+k(b-c)^2} \geq \frac34$

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Misalkan $a,b,c$ adalah bilangan riil tak negatif sehingga $a+b+c=1$. Tentukan nilai terbesar $k$ sehingga ketaksamaan berikut berlaku \[\frac{a}{1+3bc+k(b-c)^2} + \frac{b}{1+3ca+k(c-a)^2} + \frac{c}{1+3ab+k(a-b)^2} \geq \frac34.\]

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