For which $k$ does $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ hold? By generalizing this (1) and this (2) questions and performing some research

$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \frac{a+b+c}{\sqrt[3]{abc}},\hbox{ for }a,b,c>0$$

for all $0\le k<k_0\approx 11.108$.
The main goal was to prove the original inequality from (2), however, letting $a=x^3,\,b=y^3,\,c=z^3$ and clearing the denominator, the inequality becomes
$$3 k x^3 y^3 z^3 + 3 \sum\limits_{sym}x^6 y^3 z^0 - \left(3+\frac k2\right)\sum\limits_{sym} x^5 y^2 z^2\ge 0\tag{1}$$
and I'm failing to apply Muirhead's inequality.
The method from this answer works only for $k\le 3$, and even with calculus I don't think that solving system of $3$ equations like $\frac{\partial}{\partial x}$LHS(1)$=0$:
$$5 k x^3 y^2 z^2 - 9 k x y^3 z^3 + 2 k y^5 z^2 + 2 k y^2 z^5 - 18 x^4 y^3 - 18 x^4 z^3 + 30 x^3 y^2 z^2 - 9 x y^6 - 9 x z^6 + 12 y^5 z^2 + 12 y^2 z^5=0$$
may lead to something neat.)
Any help is appreciated. Thanks.
The question: what is $k_0$.
 A: Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to find a maximal $k$ for which the following  inequality is true for any positives $a$, $b$ and $c$.
$$\frac{9uv^2}{w^3}+k-3\geq \left(2+\frac{k}{3}\right)\frac{3u}{w},$$
which says that it's enough to show it for a minimal value of $v^2$.
Now, $a$, $b$ and $c$ are roots of the equation $$(x-a)(x-b)(x-c)=0$$ or
$$x^3-3ux^2+3v^2x-w^3=0$$ or
$$3v^2x=-x^3+3ux^2+w^3.$$
Id est, the line $y=3v^2x$ and the graph of $f(x)=-x^3+3ux^2+w^3$ have three common points
(maybe less of three common points  if this line is a tangent line to the graph).
We can draw a graph of $f$: $$f'(x)=-3x(x-2u),$$ which gives that $(0,w^3)$ is a minimum point and $(2u,f(2u))$ is a maximum point.
Now, we see that $v^2$ will get a minimal value, when $y=3v^2x$ would be a tangent to the graph of $f$,
which happens for equality case of two variables.
Since our inequality is homogeneous  and symmetric we can assume $b=c=1$ and $a=x^3$, which gives
$$\frac{6(x^4+x^3+x^2+2x+1)}{x^2(x+2)}\geq k,$$ which says $$k_0=\min_{x>0}\frac{6(x^4+x^3+x^2+2x+1)}{x^2(x+2)}\approx11.10864$$
Since $$\left(\frac{6(x^4+x^3+x^2+2x+1)}{x^2(x+2)}\right)'=\frac{6(x^2+x+1)(x^3+3x^2-3x-4)}{x^3(x+2)^2},$$ we see that this minimum occurs, when $x$ is a positive root of the equation: $x^3+3x^2-3x-4=0,$ which gives $$x_{min}=2\sqrt2\cos\left(\frac{1}{3}\arccos\left(-\frac{1}{4\sqrt2}\right)\right)-1.$$
A: Suppose $abc=1,$ and let $b=c=t,\,a=\frac{1}{t^2},$ the inequality become
$$k \leqslant \frac{6(t+1)(t^3+t^2+1)}{t(2t+1)} = F(t).$$
Easy to find
$$k \leqslant k_0 = F(t_0) = \frac{9\sqrt{665}}{8}\sin{\left(\frac{\pi}{6}+\frac{1}{3}\arccos{\frac{13117\sqrt{665}}{442225}}\right)}-\frac{141}{16} = 11.1086,$$
for $ \displaystyle t_0 = \frac{\sqrt 5}{2} \cos \left(\frac{\arctan(2 \sqrt{31})}{3}\right)-\frac 14 = 0 .7345.$
Finally, we will show that the inequality below is true for all $k \leqslant k_0$
$$f(a,b,c) = (a+b+c)(ab+bc+ca) + k - 3 - \left(\frac{k}{3}+2\right)(a+b+c) \geqslant 0. $$
Indeed, asumme $a = \max \{a,b,c\}$ and $t = \sqrt{bc},$ then $a \geqslant 1,$ we have
$$f(a,b,c) - f(a,t,t) = (\sqrt b - \sqrt c)^2 \left[3(a^2+ab+bc+ca)+ 6 at - k - 6\right].$$
According to the AM-GM inequality, we have
$$a^2+ab+bc+ca \geqslant 4a t,$$
so
$$3(a^2+ab+bc+ca)+ 6 at \geqslant 18at \geqslant 18 > k_0+6 \geqslant k + 6.$$
Thefore $f(a,b,c) \geqslant f(a,t,t),$ and
$$f(a,t,t)  = f\left(\frac{1}{t^2},t,t\right) = \frac{(t-1)^2}{3t^3} \left[6(t+1)(t^3+t^2+1)-t(2t+1)k\right] \geqslant 0.$$
The proof is completed.
Note. This is Ji Chen inequality.
