Another Inequality!

Applying the generalized Holder's inequality, we have:

$(a^3+3)^{1/3}(b^3+6)^{1/3}(c^3+12)^{1/3} = (a^3+1 + 2)^{1/3}(2 + b^3+4)^{1/3}(4+8+c^3)^{1/3}$ $\geq a \sqrt[3]{2} \sqrt[3]{4} + b \sqrt[3]{8} + c \sqrt[3]{2} \sqrt[3]{4} = 2(a+b+c)$

Therefore, $(a^3+3)(b^3+6)(c^3+12) \geq 8(a+b+c)^3$

For equality to hold, we must have: $\dfrac{a^3}{1^3} = \dfrac{2}{b^3} =\dfrac{4}{8}$ and $\dfrac{a^3}{2} = \dfrac{2}{4} =\dfrac{4}{c^3}$

Therefore, $a^3 = 1$ and $a^3 = \dfrac{1}{2}$ which is impossible. On the other hand, if $a=b=1, c=2$ , then

$(a^3+3)(b^3+6)(c^3+12) = 4\times 7 \times 20 = 560 < 576 = 9(a+b+c)^3$

Therefore, the largest integer $k$ would be $\boxed{8}$ .

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Note that the largest real number $k$ satisfying the inequality with the given conditions is: $k = \lambda^2$ , where: $\lambda = 7 \cos \left( \dfrac13 \cos^{-1}\left( \dfrac{89}{343} \right) \right) - \dfrac{7}{2}$

$\lambda$ is the unique positive real root to the equation $2x^3 + 21x^2 -216=0$ . Also $k \approx 8.093$ , satisfying our solution.