prove $a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b)$

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,

$$f(a)\geq f(b) = b^3-b^3-b^2c-b^3-bc^2+3b^2c+b^3+c^3-b^2c-bc^2 = $$ $$ = c^3+b^2c-2bc^2 = c(c-b)^2\geq 0.$$

Actually, a solution by AM-GM is possible. Schur is equivalent to:

$$abc\geq (-a+b+c)(a-b+c)(a+b-c).$$

In the case that all the terms of RHS is non-negative, then applying AM-GM like $$\sqrt{(-a+b+c)(a-b+c)}\leq\dfrac{-a+b+c +a-b+c}{2}=c$$ will yield the inequality.

If exactly one or all three of the terms negative, then there is nothing to prove as the product will be negative. If say, $a-b+c$ and $-a+b+c$ were negative, then their sum $2c$ is negative, which contradicts the positivity assumption, so we are done.

Let $a\geq b\geq c$.

Thus, $$\sum_{cyc}(a^3-a^2b-a^2c+abc)=\sum_{cyc}a(a-b)(a-c)\geq$$ $$\geq a(a-b)(a-c)+b(b-a)(b-c)=(a-b)^2(a+b-c)\geq0.$$