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 $\frac{4\sqrt7-7}{6}\leq p\leq\frac{65-\sqrt{130}}{42}$.
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).
Best Answer
I think, TL method does not help here.
Another way.
Let $a=x^3$, $b=y^3$ and $c=z^3$.
Thus, $x^3+y^3+z^3=3$ and we need to prove that: $$x+y+z\geq x^3y^3+x^3z^3+y^3z^3$$ or $$(x+y+z)^3(x^3+y^3+z^3)^5\geq243(x^3y^3+x^3z^3+y^3z^3)^3.$$ Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, we need to prove that $f(w^3)\geq0,$ where $$f(w^3)=u^3(9u^3-9uv^2+w^3)^5-(9v^6-9uv^2w^3+w^6)^3.$$ But, $$f'(w^3)=5u^3(9u^3-9uv^2+w^3)^4+2(9uv^2-2w^3)(9v^6-9uv^2w^3+w^6)^2\geq0,$$ which says that $f$ increases and it's enough to prove our inequality for the minimal value of $w^3$.
Now, $x$, $y$ and $z$ are positive roots of the equation $$(t-x)(t-y)(t-z)=0$$ or $$t^3-3ut^2+3v^2t-w^3=0,$$ which says that a graph of $f(t)=t^3-3ut^2+3v^2t$ and the line $g(t)=w^3$ have three common points.
But $$f'(t)=3t^2-6ut+3v^2=3(t-(u+\sqrt{u^2-v^2}))(t-(u-\sqrt{u^2-v^2})),$$ which gives $t_{min}=u+\sqrt{u^2-v^2}$ and since $f(0)=0$, we obtain two cases:
In this case $w^3$ gets a minimal value, when the line $g(t)=w^3$ is a tangent line to the graph of $f$, which says that in this case it's enough to prove our inequality for equality case of two variables.
We'll prove this for the inequality $$(x+y+z)^3(x^3+y^3+z^3)^5\geq243(x^3y^3+x^3z^3+y^3z^3)^3.$$ It's enough to assume $y=z=1$ and we need to prove that: $$(x+2)^3(x^3+2)^5\geq243(2x^3+1)^3$$ or $h(x)\geq0,$ where $$h(x)=3\ln(x+2)+5\ln(x^3+2)-3\ln(2x^3+1)-5\ln3.$$ But $$h'(x)=\frac{(x-1)(3x^5+7x^4+7x^3+6x^2-x-1)}{(x+2)(x^3+2)(2x^3+1)},$$ which gives $x_{min}=1$, $x_{max}=0.375...$ and since $h(1)=0$ it's enough to prove that $h(0)\geq0$ or $256\geq243,$ which is true.
Let $z\rightarrow0^+$ and $y=1$.
Thus, we need to prove that $$(x+1)^3(x^3+1)^5\geq243x^9,$$ which is true by AM-GM because $256>243.$