Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, by your work we need to prove that
$$\sum_{cyc}\left(a^2b+a^2c+4+2\sqrt{(a+b)(a+c)(ab+2)(ac+2)}\right)\geq\frac{27}{4}\prod_{cyc}(a+b).$$
But by AM-GM $$2\sum_{cyc}\sqrt{(a+b)(a+c)(ab+2)(ac+2)}\geq6\sqrt[3]{\prod_{cyc}((a+b)(ab+2))}$$ and it's enough to prove that
$$\sum_{cyc}(a^2b+a^2c+4a)+6\sqrt[3]{\prod_{cyc}((a+b)(ab+2))}\geq\frac{27}{4}\prod_{cyc}(a+b)$$ or
$$9uv^2-3w^3+12u^3+6\sqrt[3]{(9uv^2-w^3)\prod_{cyc}(ab+2u^2)}\geq\frac{27}{4}(9uv^2-w^3)$$ or
$$4(4u^3+3uv^2-w^3)-9(9uv^2-w^3)+8\sqrt[3]{(9uv^2-w^3)(w^6+3uw^3\cdot2u^2+3v^2\cdot4u^4+8u^6)}\geq0$$ or $f(w^3)\geq0,$ where
$$f(w^3)=(16u^3-69uv^2+5w^3)^3+512(9uv^2-w^3)(w^6+6u^3w^3+12u^4v^2+8u^6).$$
But $$f''(w^3)=-3744u^3-1134uv^2-2322w^3<0,$$
which says that $f$ is a concave function and from here $f$ gets a minimal value for an extremal value of $w^3$, which by $uvw$ says that it's enough to prove $f(w^3)\geq0$ in the following cases.
- $w^3=0$.
Let $c=0$.
Thus, $b=3-a$, where $0\leq a\leq3$ and we need to prove that
$$a^2(3-a)+(3-a)^2a+12+6\sqrt[3]{3a(3-a)(a(3-a)+2)\cdot4}\geq\frac{27}{4}\cdot3a(3-a),$$ which is true even for any real value of $a$.
- Two variables are equal.
Let $b=c=1$ after homogenization.
Thus, it's enough to prove that:
$$\left(\frac{16(a+2)^3}{27}-\frac{23(a+2)(2a+1)}{3}+5a\right)^3+$$
$$+512((a+2)(2a+1)-a)\left(a^2+\frac{2(a+2)^3a}{9}+\frac{4(a+2)^4(2a+1)}{81}+\frac{8(a+2)^6}{729}\right)\geq0$$ or
$$(a-1)^2(512a^7-1856a^6+1170456a^5+6520049a^4+13118644a^3+12182070a^2+5225276a+835921)\geq0,$$
which is obvious.