<s>[INST] prove $a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b)$ 
prove $$a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b),$$$a,b,c>0$

Obviously this is a direct consequence of  the  third degree schur's inequality.
I was wondering if this could be proved without this theorem ,or uvw  but through basic methods like AM-GM,C-S etc.
 [/INST] We write your inequality as$:$
$$\sum a(a-b)(a-c) \geqslant 0.$$
Due to symmetric$,$ by assuming $a :\neq \max\,\{a,b,c\},$ we have$:$
$$\sum a(a-b)(a-c) = a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)$$
$$=a(a-b)(a-c) +[b(b-c)(b-a)+c(c-a)(c-b)]$$
$$=a(a-b)(a-c) +(b-c)^2(b+c-a) \geqslant 0,$$
which is SOS-Schur method.
Else method is$:$
$$(a+b+c) \sum a(a-b)(a-c) = \sum a^2(a-b)(a-c) +\sum (a^3b+b^3a-2a^2b^2)$$
which is just Schur degree $4$ and AM-GM.
Schur degree $4$ is easy prove.
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