For $x,y,z \ge -1$ and $x+y+z=1$, Prove:
$$ \frac{8x}{yz}+\frac{8y}{zx}+\frac{8z}{xy}+\frac{1}{x^2y^2z^2}+9 \ge \frac{30}{xyz} $$
The difficulties of this qustion are that the constraint condition is $x,y,z \ge -1$ instead of $x,y,z \ge 0$ and it contains $x^2y^2z^2$ after multiple $x^2y^2z^2$ to both sides.
My progress is here: I considered the $pqr$ method, Let $p=x+y+z=1,q=xy+yz+zx,r=xyz$
$$
8\sum_{cyc}x^3yz+9x^2y^2z^2+1 \ge 30xyz \\
\Leftrightarrow 9r^2-2qr-22r+1 \ge 0 \qquad(*)
$$
Let $f(q)=9r^2-2qr-22r+1$, it is eazy to say that $f(q)$ is a monotonous fuction. So theoretically speaking, if we can have the extremum of $xy+yz+zx$ where $x,y,z \ge -1$ and $ x+y+z=1$, this problem will be solved. But I have no idea how to cope with it.
My second thought is to rewritten the inequality into SOS, but $x^2y^2z^2$ is 6-power which makes the computing is hard without computer.
So, is there any solution not involving computer?
I think it means that $xyz\neq0$.
Let $x=\frac{4}{3}a-1$, $y=\frac{4}{3}b-1$ and $z=\frac{4}{3}c-1$.
Thus, $a$, $b$ and $c$ are non-negatives, $a+b+c=3$ and we need to prove that: $$8\sum_{cyc}\left(\frac{4}{3}a-1\right)^2\prod_{cyc}\left(\frac{4}{3}a-1\right)+1+9\prod_{cyc}\left(\frac{4}{3}a-1\right)^2\geq30\prod_{cyc}\left(\frac{4}{3}a-1\right)$$ or $$8\prod_{cyc}(4a-3)\sum_{cyc}(4a-3)^2+243+3\prod_{cyc}(4a-3)^2\geq270\prod_{cyc}(4a-3)$$ or $$8\prod_{cyc}(4a-3)(16(a^2+b^2+c^2)-45)+243+3\prod_{cyc}(4a-3)^2\geq270\prod_{cyc}(4a-3).$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u^2=tv^2$.
Thus, $t\geq1$ and we need to prove that: $$8(64w^3-144uv^2+81u^3)(33u^2-32v^2)u+81u^6+(64w^3-144uv^2+81u^3)^2\geq90(64w^3-144uv^2+81u^3)u^3$$ or $$16w^6+4u(21u^2-34v^2)w^3+9u^2(3u^2-5v^2)^2\geq0.$$ We see that if $21u^2-34v^2\geq0$ so the inequality is true.
Thus, it's enough assume that $1\leq t\leq\frac{34}{21}.$
In another hand, if $4u^2(21u^2-34v^2)^2-144u^2(3u^2-5v^2)^2<0$ so the inequality is true and it's enough to prove our inequality for $$4u^2(21u^2-34v^2)^2-144u^2(3u^2-5v^2)^2\geq0$$ or $$4u^2(3u^2-4v^2)(39u^2-64v^2)\geq0,$$ which with $1\leq t\leq\frac{34}{21}$ gives $1\leq t\leq\frac{4}{3}$ and it's enough to prove that: $$w^3\leq\frac{-2u(21u^2-34v^2)-\sqrt{4u^2(3u^2-4v^2)(39u^2-64v^2)}}{16}$$ or $$w^3\leq\frac{34uv^2-21u^3-u\sqrt{(3u^2-4v^2)(39u^2-64v^2)}}{8}.$$ But $$\prod_{cyc}(a-b)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)\geq0$$ gives $$w^3\leq3uv^2-2u^3+2\sqrt{(u^2-v^2)^3}$$ and it's enough to prove that: $$3uv^2-2u^3+2\sqrt{(u^2-v^2)^3}\leq\frac{34uv^2-21u^3-u\sqrt{(3u^2-4v^2)(39u^2-64v^2)}}{8}$$ or $$10uv^2-5u^3-16\sqrt{(u^2-v^2)^3}\geq u\sqrt{(3u^2-4v^2)(39u^2-64v^2)}$$ or $$10-5t-16\sqrt{\frac{(t-1)^3}{t}}\geq\sqrt{(3t-4)(39t-64)}.$$ We'll prove that $$10-5t-16\sqrt{\frac{(t-1)^3}{t}}\geq0.$$ Indeed, it's $$25t(2-t)^2\geq256(t-1)^3$$ or $$231t^3-668t^2+668t-256\leq0$$ or $$(3t-4)(77t^2-120t+62)+2t-8\leq0,$$ which is true because $60^2-77\cdot62<0$.
Thus, it's enough to prove that: $$\left(10-5t-16\sqrt{\frac{(t-1)^3}{t}}\right)^2\geq(3t-4)(39t-64)$$ or $$\frac{64(t-1)^3}{t}-40(2-t)\sqrt{\frac{(t-1)^3}{t}}\geq23t^2-62t+39$$ or $$\frac{64(t-1)^2}{t}-40(2-t)\sqrt{\frac{(t-1)}{t}}\geq23t-39$$ or $$41t^2-89t+64\geq40(2-t)\sqrt{t(t-1)}$$ or since $41t^2-89t+64>0,$ $$(41t^2-89t+64)^2\geq1600(2-t)^2t(t-1)$$ or $$(9t^2+39t-64)^2\geq0$$ and we are done!
It's interesting that since $1\leq\frac{-13+5\sqrt{17}}{6}\leq\frac{3}{4},$ the equality occurs in seven different points!