<s>[INST] How prove this inequality $(a^3+b^3+c^3)(ab+bc+ac)\ge 6abc(a^2+b^2+c^2-ab-bc-ac)$ let $a,b,c$ are postive numbers, show that

$$(a^3+b^3+c^3)(ab+bc+ac)\ge 6abc(a^2+b^2+c^2-ab-bc-ac)$$

my try: let
$$a+b+c=p,ab+bc+ac=q,abc=r$$
and the
$$a^3+b^3+c^3=(a+b+c)^3-3(ab+bc+ac)(a+b+c)=p^3-3pq$$
Thank you 
 [/INST] Note that:
$$
(a^3+b^3+c^3)\geq a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ac)
$$
Therefore it is enough to prove that
$$
(a+b+c)(ab+bc+ac)\ge 6abc
$$
This is done as follows:
$$
(a+b+c)(ab+bc+ac)\ge 3(abc)^{\frac{1}{3}}\times 3(abc)^{\frac{2}{3}}=9abc \geq 6abc.
$$
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