Some thoughts.
By Holder, we have
$$\left(\sum_{\mathrm{cyc}}\sqrt{\frac{a+b}{c+ab}}\right)^2
\sum_{\mathrm{cyc}} (a + b)^2(c + ab)(c + 2)^3
\ge{} \left(\sum_{\mathrm{cyc}} (a + b)(c + 2)\right)^3.$$
It suffices to prove that
$$\left(\sum_{\mathrm{cyc}} (a + b)(c + 2)\right)^3
\ge 9 \sum_{\mathrm{cyc}} (a + b)^2(c + ab)(c + 2)^3. \tag{1} $$
This inequality is true which is verified by Mathematica. It can be proved by the pqr method.
Let $p = a + b + c, q = ab + bc + ca, r = abc$.
The condition $a + b + c + abc = 4$ becomes $p + r = 4$.
Using $r = 4- p$, (1) is written as
\begin{align*}
&208\,{q}^{3}+ \left( -72\,{p}^{2}+96\,p-2304 \right) {q}^{2}+ \left(
480\,{p}^{2}-5616\,p+15552 \right) q\\
&\quad +2384\,{p}^{3}-14184\,{p}^{2}+
34560\,p-31104
\ge 0.
\end{align*}
Omitted.
Some thoughts.
Fact 1. Let $x, y, z\ge 0$ with $x^2y^2 + y^2z^2 + z^2x^2 + \frac52 x^2 y^2 z^2 \ge \frac{11}{2}$. Then $x + y + z \ge 3$.
(The proof is given at the end.)
Let
$$x := \sqrt{\frac{ab+9}{ab+9c}}, \quad y := \sqrt{\frac{bc+9}{bc+9a}}, \quad z := \sqrt{\frac{ca+9}{ca+9b}}.$$
We have
$$x^2y^2 + y^2z^2 + z^2x^2 + \frac52 x^2 y^2 z^2 - \frac{11}{2} \ge 0. \tag{1}$$
(1) is verified by Mathematica.
By Fact 1, we have $x + y + z \ge 3$.
$\phantom{2}$
Proof of Fact 1.
Equivalently, we need to prove that, for all $x, y, z\ge 0$ with $x+y+z < 3$,
$$x^2y^2 + y^2z^2 + z^2x^2 + \frac52x^2y^2z^2 < \frac{11}{2}. \tag{A1}$$
WLOG, assume that $x + y < 2$. It suffices to prove that, for all $x, y \ge 0$ with $x + y < 2$,
$$x^2y^2 + y^2(3 - x - y)^2 + (3 - x - y)^2x^2 + \frac52x^2y^2(3 - x - y)^2 < \frac{11}{2}. \tag{A2}$$
Let $p = x + y, q = xy$. We have $0 \le p < 2$ and $p^2 \ge 4q$.
(A2) is written as
$$f(q) := -\frac{5(p-3)^2 + 2}{2}q^2 + (2p^2 - 12p + 18)q - 9p^2 - p^4 + 6p^3 + \frac{11}{2} > 0.$$
Note that $f(q)$ is concave.
Also, we have $f(0) = - 9p^2 - p^4 + 6p^3 + \frac{11}{2} > 0$
and
$$f(p^2/4) = \frac{-5p^4 + 10p^3 - 3p^2 + 44p + 44}{32}(2 - p)^2 > 0.$$
Thus, $f(q) > 0$ on $[0, p^2/4]$. We are done.
Best Answer
The desired inequality is written as $$\sum_{\mathrm{cyc}}\frac{a}{3}\sqrt{b^2 + c^2 + \sqrt{abc}^2} \ge \sqrt{3abc}. \tag{1}$$
We may use Jensen's inequality.
Let $f(x, y, z) := \sqrt{x^2 + y^2 + z^2}$. Then $f$ is convex. By Jensen's inequality, we have \begin{align*} \mathrm{LHS}_{(1)} &= \frac{a}{3}f(b, c, \sqrt{abc}) + \frac{b}{3}f(c, a, \sqrt{abc}) + \frac{c}{3}f(a, b, \sqrt{abc})\\ &\ge f\left(\frac{ab + bc + ca}{3}, \frac{ab + bc + ca}{3}, \sqrt{abc}\right)\\ &= \sqrt{\left(\frac{ab + bc + ca}{3}\right)^2 + \left(\frac{ab + bc + ca}{3}\right)^2 + abc}\\ &\ge \sqrt{3abc} \end{align*} where we use $(ab + bc + ca)^2 \ge 3(a + b + c)abc$ in the last inequality.
We are done.