Prove $8abc(a+b+c)^3\leq27(a^2+bc)(b^2+ca)(c^2+ab)$ Let $a\geq0, b\geq0,c\geq0$. Prove that:
$$8abc(a+b+c)^3\leq27(a^2+bc)(b^2+ca)(c^2+ab).$$
I tried to prove this inequality using only the inequality between the arithmetic mean and the geometric mean.
Does anyone succeed? Thanks!

Remark
$$\displaystyle\sum f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b)$$
means cyclic sum.
$$\displaystyle\prod f(a,b,c)=f(a,b,c)\cdot f(b,c,a) \cdot f(c,a,b)$$
means cyclic product.

denote
$\displaystyle f(a,b,c):=27\prod(a^2+bc)-8abc\sum a$
then expand it, we obtain
$$f(a,b,c)=27\sum{a^3b^3}+19\sum{a^4bc}-24\sum{a^3b^2c}-24\sum{a^2b^3c}+6a^2b^2c^2$$
It's easy to get an SOS form.
$$f(a,b,c)=22 \sum{a^3 b (b-c)^2}+5 \sum{a^3 (b-c)^2 c}+17 \sum{a (a-b)^2 b^2 c}+2 a b c\sum{a (a-b) (a-c)}\geq 0 $$
