Some thoughts.
By Holder inequality, we have
\begin{align*}
&\left(\sum_{\mathrm{cyc}}\frac{4a + 1}{\sqrt{4a + bc}}\right)^2 \sum_{\mathrm{cyc}} (4a + 1)(4a + bc)(ab + 6bc + 3ca + a + 4b + 2c)^3\\
\ge{}& \left(\sum_{\mathrm{cyc}} (4a + 1)(ab + 6bc + 3ca + a + 4b + 2c)\right)^3. \tag{1}
\end{align*}
It suffices to prove that
\begin{align*}
&\left(\sum_{\mathrm{cyc}} (4a + 1)(ab + 6bc + 3ca + a + 4b + 2c)\right)^3\\
\ge{}& 36\sum_{\mathrm{cyc}} (4a + 1)(4a + bc)(ab + 6bc + 3ca + a + 4b + 2c)^3. \tag{2}
\end{align*}
(2) is true which is verified by Mathematica.
Not sure where this will lead but let's give it a go!
Let $$A^2 = \frac{1}{a+1}, \quad B^2 = \frac{1}{b+1}, \quad C^2 = \frac{1}{c+1}$$
So that $A^2 + B^2 + C^2 = 1$. (like OP mentioned)
Then we simply get $$a = \frac{1-A^2}{A^2}, \quad b = \frac{1-B^2}{B^2}, \quad c = \frac{1-C^2}{C^2}$$ so
$$ab = \left(\frac{1}{A^2}-1\right)\left(\frac{1}{B^2}-1\right)$$
$$ = \frac{1}{A^2B^2}-\frac{1}{A^2}-\frac{1}{B^2}+1$$
$$\sqrt{ab-1} = \sqrt{\frac{1-(A^2+B^2)}{A^2B^2}} = \frac{\sqrt{1-(A^2+B^2)}}{AB}$$
Plugging all of this in and squaring both sides we get
$$4\left(\frac{\sqrt{1-(A^2+B^2)}}{AB}+\frac{\sqrt{1-(B^2+C^2)}}{BC}+\frac{\sqrt{1-(A^2+C^2)}}{AC}\right)^2 \leq \,\,\,...$$
$$4\left(\frac{C}{AB}+\frac{A}{BC}+\frac{B}{AC}\right)^2 \leq \left(\frac{\sqrt{1-A^2}}{A}+\frac{\sqrt{1-B^2}}{B}+\frac{\sqrt{1-C^2}}{C}\right)^2\left(\frac{\sqrt{C^2+A^2B^2}}{AB}+\frac{\sqrt{A^2+B^2C^2}}{BC}+\frac{\sqrt{B^2+A^2C^2}}{AC}\right)$$
We can add the terms in the brackets and after simplifying get
$$\frac{4}{(ABC)^2} \leq \frac{\left(BC\sqrt{1-A^2}+AC\sqrt{1-B^2}+AB\sqrt{1-C^2}\right)^2}{(ABC)^2}\left(\frac{C\sqrt{C^2+A^2B^2}+B\sqrt{B^2+A^2C^2}+A\sqrt{A^2+B^2C^2}}{ABC}\right)$$
simplifying again
$$4ABC \leq \left(BC\sqrt{1-A^2}+AC\sqrt{1-B^2}+AB\sqrt{1-C^2}\right)^2\left(C\sqrt{C^2+A^2B^2}+B\sqrt{B^2+A^2C^2}+A\sqrt{A^2+B^2C^2}\right)$$
I wasn't sure what the most efficient way from here on out was. I had made a few attempts but they didn't seem to simplify anything any further, so the following is just something that I think might help anyone else also trying to solve this.
I don't think there is much point in showing attempts, and thought it would be better to show everyone how to get back to an inequality in $a, b, c$ alone while I keep attempting at the final solution.
To get it back into a form in terms of $a,b,c$, we can divide both sides by $(ABC)^3$ to obtain
$$\frac{4}{(ABC)^2} \leq \left(\frac{\sqrt{1-A^2}}{A}+\frac{\sqrt{1-B^2}}{B}+\frac{\sqrt{1-B^2}}{B}\right)^2\left(\sqrt{\frac{C^2}{A^2B^2}+1}+\sqrt{\frac{B^2}{A^2C^2}+1}+\sqrt{\frac{A^2}{B^2C^2}+1}\right)$$
Now we can re-write everything in terms of $a, b, c$ again using the identities, and we arrive at
$$4(a+1)(b+1)(c+1) \leq \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)$$
We can simplify the RHS slightly to get
$$4abc+12 \leq \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)$$
Again I would just like to clarify that this is incomplete! If anyone spots a mistake or inconsistency please let me know.
Edit: The inequality has been solved by a different answer, and I can't seem to get to a solution using this method, so I'll say for now unless I find a way to solve it I probably will not be updating\adding to my approach here anytime soon.
Best Answer
Alternative proof.
By given condition, we can rewrite the OP as $$(ab+bc+ca)\left(5+\frac{1}{abc}\right)-5abc+23\ge 4\sum_{\mathrm{cyc}}\sqrt{5a^3+4}.$$ Or $$\frac{ab+bc+ca}{a+b+c}\left[5(a^2+b^2+c^2)+\frac{a^2+b^2+c^2}{abc}\right]-5abc+23\ge 4\sum_{\mathrm{cyc}}\sqrt{5a^3+4},$$ $$\iff\sum_{\mathrm{cyc}}{\left[\left(5a^2+\dfrac{a}{bc}\right)(ab+bc+ca)-5a^2bc+7a+8b+8c\right]}\ge 4(a+b+c)\sum_{\mathrm{cyc}}\sqrt{5a^3+4}.$$ Now, base on symmetrical principle we should try to prove $$\left(5a^2+\frac{a}{bc}\right)(ab+bc+ca)-5a^2bc+7a+8b+8c\ge 4(a+b+c)\sqrt{5a^3+4}. \tag{*}$$ Can you end it now ?
Indeed, the $(*)$ is equivalent to $$\left(5a^2+\dfrac{a}{bc}\right).\frac{a(b+c)}{a+b+c}+8\ge4\sqrt{5a^3+4}.$$ Notice that by AM-GM $$\frac{a(b+c)}{a+b+c}=\frac{1}{\dfrac{1}{a}+\dfrac{1}{b+c}}\ge \frac{2}{\dfrac{2}{a}+\dfrac{1}{\sqrt{bc}}}.$$ Hence, it is enough to prove$$\frac{5a^2+\dfrac{a}{bc}}{\dfrac{2}{a}+\dfrac{1}{\sqrt{bc}}}+4\ge2\sqrt{5a^3+4}.$$ Dividing both side by $a,$ we will prove $$ \frac{\dfrac{1}{bc}+5a}{\dfrac{1}{\sqrt{bc}}+\dfrac{2}{a}}+\dfrac{4}{a}\ge2\sqrt{5a+\dfrac{4}{a^2}}.$$ The last inequality is true by AM-GM$$ \frac{\dfrac{1}{bc}+5a}{\dfrac{1}{\sqrt{bc}}+\dfrac{2}{a}}+\dfrac{4}{a}= \frac{5a+\dfrac{4}{a^2}}{\dfrac{1}{\sqrt{bc}}+\dfrac{2}{a}}+\frac{1}{\sqrt{bc}}+\dfrac{2}{a}\ge2\sqrt{5a+\dfrac{4}{a^2}}.$$ The proof is done. Equality holds at $a=b=c=1.$