Proof.
We'll prove the following general result for $a,b,c>0,$ the constant $0\le k\le 5$ and an arbitrary negative expression $T$ represent to $a,b,c.$
$$\color{black}{\frac{\sqrt[3]{ab}+\sqrt[3]{ac}}{2\left(\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}\right)}\cdot\left((k+1)\sqrt[3]{abc}+(5-k)T\right) +4a\ge 2\sqrt{a\left(4a+k\sqrt[3]{abc}+(5-k)T\right) }.} \tag{*}$$
For easily verifying, we prove
$$\color{black}{\frac{ab+ac}{2\left(ab+bc+ca\right)}\cdot\left((k+1)abc+(5-k)T\right) +4a^3\ge 2\sqrt{a^3\left(4a^3+kabc+(5-k)T\right) }.}$$
or
$$\color{black}{\frac{b+c}{2\left(ab+bc+ca\right)}\cdot\left[(k+1)bc+\dfrac{(5-k)T}{a}\right] +4a\ge 2\sqrt{4a^2+kbc+\dfrac{(5-k)T}{a} }.} \tag{1}$$
Notice that by AM-GM $$\frac{b+c}{2\left(ab+bc+ca\right)}=\frac{1}{2a+\dfrac{2bc}{b+c}}\ge \frac{1}{2a+\sqrt{bc}}.$$Thus, it's enough to prove the stronger inequality of $(1)$ $$\frac{(k+1)bc+\dfrac{(5-k)T}{a}}{2a+\sqrt{bc}}+4a\ge 2\sqrt{4a^2+kbc+\dfrac{(5-k)T}{a} },$$which is true by AM-GM.
Indeed,
\begin{align*}
\frac{(k+1)bc+\dfrac{(5-k)T}{a}}{2a+\sqrt{bc}}+4a&=\frac{(k+1)bc+\dfrac{(5-k)T}{a}}{2a+\sqrt{bc}}+2a-\sqrt{bc}+2a+\sqrt{bc}\\&=
\frac{4a^2+kbc+\dfrac{(5-k)T}{a}}{2a+\sqrt{bc}}+2a+\sqrt{bc}\\&\ge 2\sqrt{4a^2+kbc+\dfrac{(5-k)T}{a} }.
\end{align*}
Hence, $(1)$ is proven and $(*)$ is subsequently true.
When $k=0; T=1$ we obtain $$\color{black}{\frac{\sqrt[3]{ab}+\sqrt[3]{ac}}{2\left(\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}\right)}\cdot\left(\sqrt[3]{abc}+5\right) +4a\ge 2\sqrt{a\left(4a+5\right) }.}$$
