The Contradiction method helps!
Indeed, let $\sum\limits_{cyc}\frac{1}{\sqrt{a+8b}}<1$,$a=kx$, $b=ky$ and $c=kz$ such that $k>0$ and $$\sum_{cyc}\frac{1}{\sqrt{x+8y}}=1.$$
Thus, $$\frac{1}{\sqrt{k}}\sum_{cyc}\frac{1}{\sqrt{x+8y}}<\sum_{cyc}\frac{1}{\sqrt{x+8y}},$$ which gives $k>1$ and $$3=ab+ac+bc=k^2(xy+xz+yz)>xy+xz+yz,$$ which is a contradiction because we'll prove now that $$xy+xz+yz\geq3$$ for any positives $x$, $y$ and $z$ such that $\sum\limits_{cyc}\frac{1}{\sqrt{x+8y}}=1.$
Now, let $\frac{1}{\sqrt{x+8y}}=\frac{p}{3},$ $\frac{1}{\sqrt{y+8z}}=\frac{q}{3}$ and $\frac{1}{\sqrt{z+8x}}=\frac{r}{3}.$
Thus, $$p+q+r=3,$$ $$x=\frac{\frac{1}{p^2}-\frac{8}{q^2}+\frac{64}{r^2}}{57}\geq0,$$
$$y=\frac{\frac{1}{q^2}-\frac{8}{r^2}+\frac{64}{p^2}}{57}\geq0$$ and
$$z=\frac{\frac{1}{r^2}-\frac{8}{p^2}+\frac{64}{q^2}}{57}\geq0$$ and we need to prove that:
$$\sum_{cyc}\left(\frac{1}{p^2}-\frac{8}{q^2}+\frac{64}{r^2}\right)\left(\frac{1}{q^2}-\frac{8}{r^2}+\frac{64}{p^2}\right)\geq3\cdot57^2$$ or $$\sum_{cyc}(65p^4q^2r^2-8p^4q^4)\geq171p^4q^4r^4,$$ which after substitution $p+q+r=3u$, $pq+pr+qr=3v^2$, $pqr=w^3$ gives
$$65(9u^2-6v^2)u^4w^6-8(18u^2w^6+12v^2w^6-108uv^4w^3+81v^8)u^4\geq171w^{12}$$ or $f(w^3)\geq0,$ where
$$f(w^3)=-19w^{12}+49u^6w^6-54u^2v^4w^6+96u^5v^4w^3-72u^4v^8.$$
But by Maclaurin: $$f'(w^3)=96u^5v^4+98u^6w^3-108u^2v^4w^3-76w^9>0,$$ which says that it's enough to prove $f(w^3)\geq0$ for the minimal value of $w^3$.
Now, $p$, $q$ and $r$ are positive roots of the equation:
$$(t-p)(t-q)(t-r)=0$$ or
$$t^3-3ut^2+3v^2t-w^3=0$$ or $g(t)=w^3$, where $$g(t)=t^3-3ut^2+3v^2t.$$
Let $u=constant$ and $v=constant$ and we want to move $w^3$.
During this moving should be that the equation $g(t)=w^3$ has three positive roots.
But $$g'(t)=3t^2-6ut+3v^2=3(t-t_1)(t-t_2),$$ where $t_1=u-\sqrt{u^2-v^2}$ and $t_2=u+\sqrt{u^2-v^2},$ which says $t_{max}=t_1$ and $t_{min}=t_2$.
Also, we have: $g(0)=0$ and we can draw a graph of $g$, which intersects with a line $y=w^3$ in three points or maybe, if this line is a tangent line to the graph of $g$, so they have two common points.
We see that $w^3$ gets a minimal value, when $y=w^3$ is a tangent line to a graph of $g$ in the point $(t_2,g(t_2))$. Also, we need to check, what happens for $w^3\rightarrow0^+$.
Id est, it's enough to prove $f(w^3)\geq0$ for equality case of two variables
(the case $w^3\rightarrow0^+$ is impossible because it should be $\sum\limits_{cyc}(65a^4b^2c^2-8a^4b^4)>0$).
Now, let $q=p$ and $r=3-2p$.
Thus, $$0<p<\frac{3}{2},$$ $$\frac{64}{(3-2p)^2}-\frac{7}{p^2}\geq0$$ and $$\frac{65}{p^2}-\frac{8}{(3-2p)^2}\geq0,$$ which gives
$$\frac{4\sqrt7-7}{6}\leq p\leq\frac{65-\sqrt{130}}{42}$$ and we need to prove that
$$130p^6(3-2p)^2+65p^4(3-2p)^4-8(p^8+2p^4(3-2p)^4)\geq171p^8(3-2p)^4$$ or
$$(p-1)^2(441-294p+277p^2+152p^3-1368p^4+1216p^5-304p^6)\geq0,$$ which is true for $\frac{4\sqrt7-7}{6}\leq p\leq\frac{65-\sqrt{130}}{42}$.
Some thoughts.
Remarks: Dragon boy's answer using Ji Chen's lemma is nice.
Holder also works.
I got the same form of Holder inequality independently.
Actually, I used the same idea to obtain the form of Holder inequality here.
We use the pqr method.
Let $p = a + b + c, q = ab + bc + ca = 1, r = abc$.
The desired inequality is written as
\begin{align*}
&\left( 80\,p\sqrt {6}+66\,\sqrt {6}-230\,p-156 \right) {r}^{2}\\
&\qquad +
\left( 2\,{p}^{4}\sqrt {6}-6\,{p}^{3}\sqrt {6}+43\,{p}^{4}+67\,{p}^{2
}\sqrt {6}+21\,{p}^{3}-64\,p\sqrt {6}\right.\\
&\qquad\qquad \left. -22\,{p}^{2}-220\,\sqrt {6}+124\,
p+520 \right) r\\
&\qquad +8\,{p}^{5}\sqrt {6}+8\,{p}^{6}+{p}^{4}\sqrt {6}+{p}^{5
}-73\,{p}^{3}\sqrt {6}-46\,{p}^{4}-36\,{p}^{2}\sqrt {6}\\
&\qquad\qquad +68\,{p}^{3}+
156\,p\sqrt {6}+156\,{p}^{2}+144\,\sqrt {6}-312\,p-352\\
&\ge 0.
\end{align*}
It is quadratic in $r$. So it is tractable.
Best Answer
I hope the following will help.
By AM-GM $$\sqrt{2(a+bc)}\le a+bc+\frac{1}{2},$$hence we will prove$$\frac{1}{2a+2bc+1}+\frac{1}{2b+2ca+1}+\frac{1}{2c+2ab+1}\ge 1.$$ The last inequality was here.