Prove that $\sum\limits_{cyc}\frac{a^2}{a^2+bc}+\frac{(a+b+c)^3+9abc}{\prod\limits_{cyc}(a+b)}\geq6.$ There is the following anhduy98's problem.
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}+\frac{(a+b+c)^3+9abc}{(a+b)(a+c)(b+c)}\geq6.$$
I solved this problem by BW(about BW see here: https://artofproblemsolving.com/community/c6h522084)
and I am looking for an alternative solution.
The equality occurs also for $c=0$ and $a=b$.
I tried $uvw$(about $uvw$ see here https://artofproblemsolving.com/community/c6h278791).
It gives $$4w^9+(22u^3-39uv^2)w^6+(27u^6-99u^4v^2+81u^2v^4+13v^6)w^3+9uv^6(3u^2-4v^2)\geq0$$  and I did not get solution for variations of $u$, of $v^2$ and of $w^3$.
Also, I tried $SOS$ and I did not get a solution.
Thank you!
PS. The River Li's solution we can write in the following form.
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}+\frac{(a+b+c)^3+9abc}{(a+b)(a+c)(b+c)}-6=$$
$$=\frac{\sum\limits_{cyc}\left(4a^2b+4a^2c-\frac{1}{3}abc\right)\prod\limits_{cyc}(a-b)^2+\left(\sum\limits_{cyc}(2a^3-a^2b-a^2c)\right)^2abc}{4\prod\limits_{cyc}(a+b)\prod\limits_{cyc}(a^2+bc)}\geq0$$

Remarks: The pqr method works well, furthermore, it leads to a SOS solution directly.
Let $p = a + b + c, ~ q = ab + bc + ca, ~ r = abc$.
We need to prove that $F(p, q, r) \ge 0$ where
\begin{align*}
 F(p, q, r) &= -4\,p{q}^{4}+ \left( {p}^{3}+13\,r \right) {q}^{3}+27\,{p}^{2}r{q}^{2} \\
 &\quad
 - \left( 11\,{p}^{4}r + 117\,p{r}^{2} \right) q+{p}^{6}r+22\,{p}^{3}{r}
 ^{2}+108\,{r}^{3}.
\end{align*}
Note that
$$\Delta = -4\,{q}^{3}+{p}^{2}{q}^{2}+18\,prq-4\,{p}^{3}r-27\,{r}^{2} = (a - b)^2(b - c)^2(c - a)^2 \ge 0.$$
We have
$$F - (pq - 13r/4)\Delta = \frac{1}{4}r(2p^3 - 7pq + 9r)^2.$$
Since $pq \ge 9r \ge 13r/4$, we have $F \ge 0$. We are done.
