EDIT: There is some problem in the original proof below. Specifically $p^3r\geqslant q^3$ does not necessarily hold. I found this inequality accidently at that time but did not check the correctness. So, sorry for my carelessness.
Here is a new proof (still complicated, though).
Our goal is to prove
$$
\sum\sqrt{5a^2+4ab}\geqslant9=3\sqrt{3(ab+bc+ca)}.
$$
We can prove
$$
\sum\sqrt{\frac{5a^2+4ab}{3(ab+bc+ca)}}\geqslant3.
$$
We can simplify the square root operation, by the following equality:
$$
\sqrt{t}\geqslant\frac{6t^3+20t^2+6t}{t^3+15t^2+15t+1},
$$
which can be derived from $(\sqrt t-1)^6\geqslant0\iff t^3+15t^2+15t+1\geqslant\sqrt t(6t^2+20t+6)$.
Therefore
$$
\sqrt t-1\geqslant\frac{(t - 1) (5 t^2 + 10 t + 1)}{(t + 1) (t^2 + 14 t + 1)}=:f(t).
$$
Let
$$
u=\frac{5a^2+4bc}{3(ab+bc+ca)},\, v=\frac{5b^2+4bc}{3(ab+bc+ca)},\, w=\frac{5c^2+4ca}{3(ab+bc+ca)}.
$$
It can be proved
$$
f(u)+f(v)+f(w)\geqslant0.
$$
It can be verified with the help of Mathematica. We can assume without loss of generality $c=\min(a,b,c)$. We need to consider both the case $c\leqslant a\leqslant b$ and $c\leqslant b\leqslant a$. The first case is simplier than the second. For the second case, the minimum is achieved only if $c=0$, so by the homogeneity we just need to consider $(a,b,c)=(a,1,0)$. The rest is just to verify the positiveness for all $a$ in such case.
The proof is done.
PS: If the estimation of $\sqrt t$ is based on $(\sqrt t-1)^4\geqslant0$ instead of $(\sqrt t-1)^6$, it seems the positiveness cannot be guaranteed in the case $c\leqslant b\leqslant a$.
Original answer
This is NOT a nice solution so only the idea is sketched.
You have proved $\sum\sqrt{5a^2+4ab}\cdot\sqrt{5b^2+4bc}\geqslant 15+4\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)$. Hence
$$
\sum\sqrt{5a^2+4ab}\cdot\sqrt{5b^2+4bc}\geqslant15+12(abc)^{2/3}.
$$
Then, in order to prove the original inequality, we just need to prove
$$
\begin{gathered}
\vphantom{\sum}\\
\impliedby\vphantom{a^2} \\
\impliedby\vphantom{a^2}
\end{gathered}
\begin{gathered}
5(a^2+b^2+c^2)+2\sum \sqrt{5a^2+4ab}\cdot\sqrt{5b^2+4bc}\geqslant 69\\
5(a^2+b^2+c^2)+24(abc)^{2/3}\geqslant39\\
5(a^2+b^2+c^2)+24(abc)^{2/3}\geqslant13(ab+bc+ca).
\end{gathered}
$$
This is true but in a short time I haven't found an elegant proof.
There is an ugly proof available. Let $a=x^3$, $b=y^3$, $c=z^3$, $p=x+y+z$, $q=xy+yz+xz$, $r=xyz$. We can prove
$$
5\sum x^6+24(xyz)^2\geqslant 13\sum x^3y^3,
$$
using the facts that
$$
\begin{gathered}
\sum x^6= p^6−6p^4q+6p^3r+9p^2q^2−2q^3−12pqr+3r^2,\\
\sum x^3y^3=q^3-3pqr+3r^2,\\
p^3r\geqslant q^3,\\
p^3r\geqslant 3pqr.
\end{gathered}
$$
After simplification, the only thing left is
$$
5p^6+45p^2q^2\geqslant5p^2\cdot 6p^2q,
$$
which is obvious.
