<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] Easiest proof is actually the proof of Schur. Alternatively, you can even prove it using simple calculus.
Assume $a\geq b\geq c.$ Consider:
$$f(x) = x^3-x^2(b+c) - x(b^2+c^2) +3xbc+b^3+c^3-bc(b+c),\,\, x\geq b$$
Then, $f'(x) = 3x^2-2x(b+c) - b^2-c^2+3bc$ and $f''(x) = 6x - 2b-2c\geq 0.$
So $f'$ is increasing on $[b,\infty)$ and as such:
$$f'(x)\geq f'(b) = 3b^2 - 2b^2-2bc-b^2-c^2+3bc = c(b-c)\geq 0$$
which means $f$ is increasing on its domain. Finally,
$$f(a)\geq f(b) = b^3-b^3-b^2c-b^3-bc^2+3b^2c+b^3+c^3-b^2c-bc^2 = $$
$$ = c^3+b^2c-2bc^2 = c(c-b)^2\geq 0.$$
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