<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] Suppose $a \geqslant b \geqslant c,$ we get
$$a^3+b^3+c^3+3abc - \sum ab(a+b)$$
$$=(c^3+b^2c+ca^2-abc-ac^2-bc^2)+(a^3+b^3+4abc-a^2b-ab^2-2ca^2-2b^2c)$$
$$=(a^2+b^2+c^2-ab-bc-ca)c+(a+b-2c)(a-b)^2 \geqslant 0.$$
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