Proof.
By Holder, we will prove $$\frac{(a^2+b^2+c^2+9)^3}{\sum\limits_{cyc}{(a^2+3)^3(2b^2+5bc+2c^2)}
}\ge 1.$$
After homogenization, it is
$$(ab+bc+ca)\left[(a+b+c)^2+ab+bc+ca\right]^3 \ge 3\sum_{\mathrm{cyc}}{(a+b)^3(a+c)^3(2b^2+5bc+2c^2)}.\tag{*}$$Equality holds at $(t,t,t), t>0$ or $abc=0.$
Let $a+b+c=3u; ab+bc+ca=3v^2; abc=w^3$. $(*)$ turns out $$f(w^3)=A(u,v^2)w^6+B(u,v^2)w^3+C(u,v^2)+27v^2(3u^2+v^2)^3\ge 0,\tag{**}$$where $f''(w^3)<0$ and by $uvw$ we need to prove the problem is true when $w^3$ achieve extremal value. Consider two cases:
- $w^3=0.$
- Two equal variables.
The desired result follows.
Indeed, we can demonstrate $RHS_{(*)}$ as
\begin{align*}
&2\left[(a+b)(b+c)(c+a)\right]^2.\sum_{cyc}(a+b)(a+c)+\sum_{cyc}bc(a+b)^3(a+c)^3\\&=2(9uv^2-w^3)^2(9u^2+3v^2)+w^3(a^5+b^5+c^5)+81v^8+9v^2w^3(a^3+b^3+c^3)+81uv^4w^3,
\end{align*}
Notice that: $a^3+b^3+c^3=27u^3-27uv^2+3w^3; a^5 + b^5 + c^5 = 243u^5-405u^3v^2+135uv^4+5(9u^2-3v^2)w^3.$
Now, with the form $(**)$ we obtain $$\color{blue}{A=-\left[2(9u^2+3v^2)+5(9u^2-3v^2)+27v^2 \right]=-\left(63u^2+18v^2\right)<0.}$$
Hence, $f(r)$ is concave and we prove in two cases.
- In case $w^3=0,$ $(*)$ becomes $$bc(b^2+3bc+c^2)^3\ge 3\left[(bc)^3(2b^2+5bc+2c^2)+2c^2b^3(b+c)^3+2b^2c^3(b+c)^3\right],$$
$$\iff (b^2+3bc+c^2)^3\ge 3\left[(bc)^2(2b^2+5bc+2c^2)+2cb^2(b+c)^3+2bc^2(b+c)^3\right]. \tag{1}$$
If $bc=0,$ then $(1)$ is true.
If $bc\neq 0,$ then let $b=1$ and $(1)$ turns out $$(1+3c+c^2)^3\ge 3\left[c^2(2+5c+2c^2)+2c(1+c)^3+2c^2(1+c)^3\right],$$
$$\iff c^6 + 3 c^5 - 6 c^3 + 3 c + 1\ge 0,$$
$$\iff (c^3-1)^2+c[3(c^2-1)^2+2c^2]\ge 0.$$
- In case two equal variables
Since this inequality is homogeneous and for $b=c=0$ it's obviously true, we can assume $b=c=1$, which gives $$(2a+1)\left[(a+2)^2+2a+1\right]^3 \ge 3\left[9(a+1)^6+16(a+1)^3(2a^2+5a+2)\right],$$
$$\iff 2 a^7 + 10 a^6 + 6 a^5 - 18 a^4 - 18 a^3 + 6 a^2 + 10 a + 2 \ge 0,$$
$$\iff 2(a-1)^2(a+1)^3(a^2+4a+1)\ge 0.$$
I hope to see nice proof for $(*).$
Generally, for all $n \ge \dfrac{10-3\sqrt{3}}{2},$ the following inequality
$$\boxed{\dfrac{1}{\sqrt{a^2+nab+b^2}}+\dfrac{1}{\sqrt{b^2+nbc+c^2}}+\dfrac{1}{\sqrt{c^2+nca+a^2}} \ge 3\sqrt{\dfrac{3}{(n+2)(ab+bc+ca)}},}$$
holds for $a,b,c\ge 0: ab+bc+ca>0.$
At $\color{blue}{n=\dfrac{10-3\sqrt{3}}{2},}$ equality holds also at $(0,t,t)$ ($t>0$)
