<s>[INST] Inequality : $(a^3+3)(b^3+6)(c^3+12)>k(a+b+c)^3$ Find the largest integer $k$ such that 
$$(a^3+3)(b^3+6)(c^3+12)>k(a+b+c)^3$$
for all positive real numbers $a, b, c$.
My attempt :
By Holder inequality, $(a^3+1+2)(2+b^3+4)(4+8+c^3)\geq(\sqrt[3]{8}a+\sqrt[3]{8}b+\sqrt[3]{8}c)^3$
$\Leftrightarrow(a^3+3)(b^3+6)(c^3+12)\geq8(a+b+c)^3$
There is no equality hold.
so $(a^3+3)(b^3+6)(c^3+12)(a+b+c)^3>8(a+b+c)^3$ 
 [/INST] The equality should be for $(a,1,\sqrt[3]2)||(\sqrt[3]2,b,\sqrt[4]4)||(\sqrt[3]4,2,c),$ which is impossible.
Let $a_1+a_2=3$, $b_1+b_2=6$, $c_1+c_2=12$, $b_1c_1=a_1c_2=a_2b_2$, where $a_i\geq0$, $b_i\geq0$ and $c_i\geq0$.
Thus, for the equality occurring after using Holder we need 
 $$\left(a,\sqrt[3]{a_1},\sqrt[3]{a_2}\right)||\left(\sqrt[3]{b_1},b,\sqrt[3]{b_2}\right)||\left(\sqrt[3]{c_1},\sqrt[3]{c_2},c\right),$$
which is very ugly:
$k_{max}=8.093...$ for $(a,b,c)=(0.904...,1.368...,2.297...)$,
but it says $8$ is a maximum possible integer value, for which our inequality is true 
and $(0.904,1.368,2.297)$ is a counterexample for all integer $k>8$. 
By the way, there is a counterexample, which is a bit of better: $$(a,b,c)=(1,2,2).$$
In the exam you apply your proof for $k=8$ and for $k\geq9$ you write:
The equality $$(a^3+3)(b^3+6)(c^3+12)\geq k(a+b+c)^3$$
is wrong. $(1,2,2)$ is the counterexample.
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