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.

Another way.
Let $a\geq b\geq c$.
Thus, $$\sum_{cyc}(a^3-a^2b-a^2c+abc)=\sum_{cyc}(a^3-abc-a^2b-a^2c+2abc)=$$
$$=\sum_{cyc}(a-b)^2\left(\frac{a+b+c}{2}-c\right)=\frac{1}{2}\sum_{cyc}(a-b)^2(a+b-c)\geq$$
$$\geq\frac{1}{2}((a-c)^2(a+c-b)+(b-c)^2(b+c-a))\geq$$
$$\geq\frac{1}{2}((b-c)^2(a-b)+(b-c)^2(b-a))=0.$$
