Problem. Given three numbers $a, b, c\geqq 0$ and $k= constant$ so that $- \dfrac{2}{11}\leqq k\leqq 0$. Prove that : $$(\!k+ 1\!)^{6}(\!a+ b+ c\!)^{2}(\!ab+ bc+ ca\!)^{2}\!- 81(ka+ b)(kb+ a)(kb+ c)(kc+ b)(kc+ a)(ka+ c)\!\geqq 0$$
Remark. By using discriminant, I determined the value of $k$ so that the inequality holds for $a, b, c\geqq 0$, thus I gave $a= b= 1$ and $c\rightarrow 0^{+}$, and I also have the coefficient of $a^{4}$ must be positive. I make it involving a pretty good inequality as the following one for $a, b, c\geqq 0$ and $k= constant$ : $$3\sum \sqrt{(a+ kb)(a+ kc)}\leqq (k+ 1)\left ( (a+ b+ c)+ 2\sqrt{3(ab+ bc+ ca)} \right )\,\therefore\,k= 2\,only\,true\,!$$ This one that Ji Chen's Symmetric Function Theorem applies. Return the OP, we may use $uvw$ here !
$$\prod_{cyc}(ka+b)(kb+a)=\prod_{cyc}(k(a^2+b^2)+(k+1)ab)=$$ $$=\sum_{cyc}(k^3(a^4b^2+a^4c^2)+(k^3+k^2)a^3b^3+(k^3+k^2)a^4bc)+$$ $$+abc\sum_{cyc}\left((2k^3+3k^2+k)(a^2b+a^2c)+\left(k^3+k^2+k+\frac{1}{3}\right)abc\right).$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, the coefficient before $w^6$ s equal to $$-3k^3+3(k^3+k^2)+3(k^3+k^2)-3(2k^3+3k^2+k)+3\left(k^3+k^2+k+\frac{1}{3}\right)=1,$$ which says that our inequality is equivalent to $f(w^3)\geq0,$ where $f$ is a concave function.
But the concave function gets a minimal value for the extreme value of $w^3$, which happens in the following cases:
For $b=c=0$ our inequality is an equality for all $k$.
Let $b=1$ and $c=0$.
We obtain: $$(k+1)^6(a+1)^2a^2\geq81k^2(ka+1)(a+k)a^2$$ or $$((k+1)^6-81k^3)a^2+(2(k+1)^6-81k^2(k+1))a+(k+1)^6-81k^3\geq0$$ and since $(k+1)^6-81k^3>0$, for $1(k+1)^6-81k^2(k+1)\geq0$ the equality is true.
But for $2(k+1)^6-81k^2(k+1)<0$ we need $$(2(k+1)^6-81k^2(k+1))^2-4((k+1)^6-81k^3)^2\leq0$$ or $$(1-k)k^2(4k^6+24k^5+60k^4-163k^3-21k^2-24k+4)\geq0,$$ which is wrong for $k=-\frac{2}{11}$ and from here you can make a counterexample.
The case $b=c=1$ is not relevant already.