Maybe the following will help.
By Holder $$\sum\limits_{cyc}\frac{1}{\sqrt{a^2+b^2}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{1}{\sqrt{a^2+b^2}}\right)^2\sum\limits_{cyc}(a^2+b^2)(ab+\sqrt2c^2)^3}{\sum\limits_{cyc}(a^2+b^2)(ab+\sqrt2c^2)^3}}\geq\sqrt{\frac{\left(\sum\limits_{cyc}(\sqrt2a^2+ab)\right)^3}{\sum\limits_{cyc}(a^2+b^2)(ab+\sqrt2c^2)^3}}$$ and it's enough to prove that:
$$8\left(\sum\limits_{cyc}(\sqrt2a^2+ab)\right)^3\geq(1+2\sqrt2)^2\sum\limits_{cyc}(a^2+b^2)(ab+\sqrt2c^2)^3,$$ which saves the case of an equality occurring and it's true for $b=c=0$ and $a=4$.
I checked also that it's true for $b=a$ and $c=\frac{4-2a}{1+a^2},$ where $0\leq a\leq2$.
The term $(\sqrt2c^2+ab)^3$ we can get by the following way.
Note by Holder $$\left(\sum_{cyc}\frac{1}{\sqrt{a^2+b^2}}\right)^2\sum_{cyc}(a^2+b^2)(a^2+b^2+kc^2+mab+nac+nbc)^3\geq$$
$$\geq\left(\sum_{cyc}((2+k)a^2+(m+2n)ab)\right)^3.$$
The equality occurs, when $$\left(\frac{1}{\sqrt{a^2+b^2}},\frac{1}{\sqrt{b^2+c^2}},\frac{1}{\sqrt{c^2+a^2}}\right)||$$
$$||\left((a^2+b^2)(a^2+b^2+kc^2+mab+nac+nbc)^3,(b^2+c^2)(b^2+c^2+ka^2+mbc+nab+nac)^3,(c^2+a^2)(c^2+a^2+kb^2+mac+nab+nbc)^3\right),$$ which gives
$$\sqrt{a^2+b^2}(a^2+b^2+kc^2+mab+nac+nbc)=\sqrt{b^2+c^2}(b^2+c^2+ka^2+mbc+nab+nac)=\sqrt{c^2+a^2}(c^2+a^2+kb^2+mac+nab+nbc).$$
For $b=c=0$ and $a=4$ the first equation gives $$4\cdot16=0\cdot16k,$$ which is possible, when coefficients before $a^2$ and $b^2$ in the expression $(a^2+b^2+kc^2+mab+nac+nbc)^3$ are equal to $0$ and for the equality case we obtain:
$$\sqrt{a^2+b^2}(c^2+mab+nac+nbc)=\sqrt{b^2+c^2}(a^2+mbc+nab+nac)=\sqrt{c^2+a^2}(b^2+mac+nab+nbc),$$ which for $a=b=2$ and $c=0$ gives:
$$2\sqrt2\cdot4m=2(4+4n),$$ which for $n=0$(because with $0$ we obtain something simple) gives $m=\frac{1}{\sqrt2}$ and we got an expression $$(\sqrt2c^2+ab)^3,$$ for which Holder saves the case of an equality occurring.
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 $\frac{4\sqrt7-7}{6}\leq p\leq\frac{65-\sqrt{130}}{42}$.
Best Answer
We use the isolated fudging.
It suffices to prove that, for all $a, b, c > 0$, $$\frac{2}{3\sqrt 6}\sqrt{\frac{ab + 2(a + b + c)^2/9}{ab + c(a + b + c)/3}}\ge \frac{12(a^2 + b^2) + 32ab + 35c(a + b)}{24(a^2 + b^2 + c^2) + 102(ab + bc + ca)}. \tag{1}$$ (Summing cyclically on (1), the desired result follows.)
(1) can be proved by Buffalo Way (BW).