Partial simple answer :
Let :
$$f\left(x\right)=2\sqrt{3}\left(\frac{1}{\sqrt{e^{-x}+2}}\right)$$
Then :
$$f''(x)\leq 0,\forall x\in[-\ln(4),\infty)$$
Then using Jensen's inequality we have $x\in[\frac{1}{4}\max\left(a,b,c\right),\infty]$ and $b\to b/x,a\to a/x,c\to c/x$:
$$2\sqrt{3}\left(\frac{1}{\sqrt{\frac{a}{x}+2}}+\frac{1}{\sqrt{\frac{b}{x}+2}}+\frac{1}{\sqrt{\frac{c}{x}+2}}\right)\leq 3g\left(\frac{x}{\left(abc\right)^{\frac{1}{3}}}\right),g\left(x\right)=2\sqrt{3}\left(\frac{1}{\sqrt{1/x+2}}\right)$$
Then I think you can finish that .
Edit for more details :
$$3*2\sqrt{3}\left(\frac{1}{\sqrt{\frac{\left(abc\right)^{\frac{1}{3}}}{x}+2}}\right)\leq 5+\frac{x}{\left(abc\right)^{\frac{1}{3}}}$$
Setting : $\frac{x}{\left(abc\right)^{\frac{1}{3}}}=1/y$ so :
$$3*2\sqrt{3}\left(\frac{1}{\sqrt{y+2}}\right)\leq 5+1/y$$
It's a second degree polynomial so think to push on both side the square .
Remark to have a complete proof :
We can use the intermediate inequality for $x\ge 0$ :
$$\frac{2\sqrt{3}}{\sqrt{x+2}}\leq f\left(x\right)=\frac{6}{\sqrt{x}+2}$$
Then use Jensen's inequality for $a,b\in[1,\infty)$ we have to show :
$$2f\left(\sqrt{ab}\right)+f\left(c\right)-5-\frac{1}{\left(abc\right)^{\frac{1}{3}}}\leq 0$$
Then we plugg $ab=u$ .
The inequality now is less harder .