let $a,b,c\in R$,and such $$ab+bc+ac=3$$ show that $$4(a^2+b^2+c^2)+9a^2b^2c^2\ge 21$$
if this equlity condition is $a,b,c$ be postive numbers,and I can use $pqr$ to prove it.
But this is $a,b,c$ are real numbers
let $a+b+c=p,ab+bc+ac=q=3,abc=r$ so $$a^2+b^2+c^2=p^2-6$$ then $$4(a^2+b^2+c^2)+9a^2b^2c^2=4p^2-24+9r^2$$ since $a,b,c\in R$,so I can't use this important schur inequality： $$9r\ge 4pq-p^3,a,b,c>0$$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u^2=tv^2$.
Hence, our inequality is equivalent to $w^6\geq v^4(5v^2-4u^2)$. Since, the inequality does not depend on substitution $a\rightarrow-a$, $b\rightarrow-b$ and $c\rightarrow-c$, we can assume $u\geq0$. Easy to show that $(a-b)^2(a-c)^2(b-c)^2\geq0\Leftrightarrow3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6$ or
$3uv^2-2u^3-2\sqrt{(u^2-v^2)^3}\leq w^3\leq3uv^2-2u^3+2\sqrt{(u^2-v^2)^3}$.
For $t\geq\frac{5}{4}$ the inequality is obviously true.
But for $1\leq t\leq\frac{5}{4}$ it remains to prove that $2\sqrt{(u^2-v^2)^3}+v^2\sqrt{5v^2-4u^2)}\leq3uv^2-2u^3$ or
$u^2(3v^2-2u^2)^2\geq\left(2\sqrt{(u^2-v^2)^3}+v^2\sqrt{5v^2-4u^2)}\right)^2$ or
$t(3-2t)^2\geq\left(2\sqrt{(t-1)^3}+\sqrt{5-4t}\right)^2$ or
$t-1\geq4\sqrt{(t-1)^3(5-4t)}$ or
$(t-1)^2(8t-9)^2\geq0$. Done!