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.
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.
Best Answer
Easy to see that for $c\rightarrow0^+$ or for $c\rightarrow+\infty$ the inequality holds.
Let $$f(a,b,c)=\frac{1}{\sqrt[3]{abc}}+5-2\sqrt3\sum_{cyc}\frac{1}{\sqrt{a+2}}.$$ Thus, in the inside minimum point should be $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0.$$ Let in this point $a\neq b$, $a\neq c$ and $b\neq c$.
Thus, $$\frac{1}{3\sqrt{a^4bc}}-\frac{\sqrt3}{\sqrt{(a+2)^3}}=\frac{1}{3\sqrt{ab^4c}}-\frac{\sqrt3}{\sqrt{(b+2)^3}}=\frac{1}{3\sqrt{abc^4}}-\frac{\sqrt3}{\sqrt{(c+2)^3}}=0,$$ which gives $$\frac{1}{3\sqrt3\sqrt[3]{abc}}=\frac{a}{\sqrt{(a+2)^3}}=\frac{b}{\sqrt{(b+2)^3}}=\frac{c}{\sqrt{(c+2)^3}}$$ and from here $$\frac{a^2}{(a+2)^3}=\frac{b^2}{(b+2)^3}$$ or $$(a-b)(a^2b^2-12ab-8a-8b)=0,$$ which by AM-GM gives $$a^2b^2=12ab+8a+8b\geq12ab+16\sqrt{ab},$$ which gives $ab\geq16$ and by the same way $ac\geq16$ and $bc\geq16.$
Also, we have $$a^2b^2-a^2c^2=12ab+8a+8b-12ac-8a-8c$$ or $$(b-c)(a^2(b+c)-12a-8)=0$$ or $$12a+8=a^2(b+c).$$ By the same way we obtain also $$12b+8=b^2(a+c)$$ and from here $$(a-b)(ab+ac+bc-12)=0,$$ which gives $$ab+ac+bc=12,$$ which is a contradiction because $$12=ab+ac+bc\geq3\cdot16\geq48.$$ Thus, in the minimum point $(a,b,c)$ two numbers should be equal.
Let $b=a$ and $c\neq a$.
Thus, since $$(a-c)(a^2c^2-12ac-8a-8c)=0,$$ we obtain $$a^2c^2-12ac-8a-8c=0$$ or $$a^2c^2-4(3a+2)c-8a=0$$ or $$c=\frac{2(3a+2)+\sqrt{4(3a+2)^2+8a^3}}{a^2}$$ or $$c=\frac{2(3a+2)+2(a+2)\sqrt{2a+1}}{a^2}$$ and since $$\frac{1}{\sqrt[3]{a^2c}}=\frac{3\sqrt3a}{\sqrt{(a+2)^3}},$$ it's enough to prove in this case that: $$\frac{3\sqrt3a}{\sqrt{(a+2)^3}}+5\geq2\sqrt3\left(\frac{2}{\sqrt{a+2}}+\frac{1}{\sqrt{\frac{2(3a+2)+2(a+2)\sqrt{2a+1}}{a^2}+2}}\right)$$ or $$\frac{3\sqrt3a}{a+2}+5\sqrt{a+2}\geq2\sqrt3\left(2+\frac{a}{\sqrt{2(a+1+\sqrt{2a+1})}}\right)$$ or $$\frac{3\sqrt3a}{a+2}+5\sqrt{a+2}\geq2\sqrt3\left(2+\frac{a}{\sqrt{2a+1}+1}\right)$$ or $$\frac{3\sqrt3a}{a+2}+5\sqrt{a+2}\geq\sqrt3\left(3+\sqrt{2a+1}\right)$$ or $$5\sqrt{a+2}\geq\sqrt3\left(\sqrt{2a+1}+\frac{6}{a+2}\right)$$ or $$19a^3+123a^2+264a+80\geq36(a+2)\sqrt{2a+1}$$ and since $$\sqrt{2a+1}\leq a+1,$$ it's enough to prove here that:$$19a^3+123a^2+264a+80\geq36(a+2)(a+1),$$ which is obvious.
For $a=c$ we need to prove that: $$\frac{1}{a}+5\geq\frac{6\sqrt3}{\sqrt{a+2}}$$ or $$(5a+1)^2(a+2)\geq108a^2,$$ which is true by AM-GM: $$(5a+1)^2(a+2)\geq\left(6\sqrt[6]{a^5}\right)^2\cdot3\sqrt[3]a=108a^2$$ and we are done!