Problem :
Let $a,b,c>0$ then we have :
$$\left[4-\frac{3\left(\sqrt{ac}+\sqrt{bc}+\sqrt{ab}\right)}{a+b+c}\right]^{\frac{3}{2}}abc-\left(\frac{16}{9\sqrt{3}}\right)^{\frac{1}{2}}\left(\frac{a+b+c}{3}\right)^{3}< 0$$
My Attempt :
We use a Theorem A :
Theorem A :
Let $x_1,x_2,x_3,y_1,y_2,y_3\in(0,\infty)$ and $u>0$ a real then if we have :
$$x_1^u+x_2^u+x_3^u\geq y_1^u+y_2^u+y_3^u$$
And for $i\neq j,1\leq i\leq 3,1\leq j\leq 3$:
$$|x_i^u-x_j^u|\leq |y_i^u-y_j^u|$$
Then we have :
$$x_1x_2x_3\geq y_1y_2y_3$$
To use the theorem A we set :
$$y_1=\left[4-\frac{3\left(\sqrt{ac}+\sqrt{bc}+\sqrt{ab}\right)}{a+b+c}\right]^{\frac{1}{2}}\sqrt{ab}$$
$$y_2=\left[4-\frac{3\left(\sqrt{ac}+\sqrt{bc}+\sqrt{ab}\right)}{a+b+c}\right]^{\frac{1}{2}}\sqrt{bc}$$
$$y_3=\left[4-\frac{3\left(\sqrt{ac}+\sqrt{bc}+\sqrt{ab}\right)}{a+b+c}\right]^{\frac{1}{2}}\sqrt{ca}$$
And :
$$x_2=x_3=x_1=\left(\frac{16}{9\sqrt{3}}\right)^{\frac{1}{6}}\left(\frac{a+b+c}{3}\right)$$
With the choose of $x_1,x_2,x_3$ we reach one of the constraint .
Here $u=0.00625$.
So we have to show after homogeneization:
$$\left(4-\frac{3\left(\sqrt{ab}+\sqrt{b}+\sqrt{a}\right)}{b+a+1}\right)^{\frac{u}{2}}\left(b^{\frac{1}{2}u}+\left(ba\right)^{\frac{1}{2}u}+a^{\frac{1}{2}u}\right)-\frac{3\left(\frac{16}{9\sqrt{3}}\right)^{\frac{u}{6}}\left(1+a+b\right)^{u}}{3^{u}}\leq 0$$
Or :
$$\left(4-\frac{3\left(\sqrt{ab}+\sqrt{b}+\sqrt{a}\right)}{b+a+1}\right)^{\frac{u}{2}}\left(b^{\frac{1}{2}u}+\left(ba\right)^{\frac{1}{2}u}+a^{\frac{1}{2}u}\right)\leq \frac{3\left(\frac{16}{9\sqrt{3}}\right)^{\frac{u}{6}}\left(1+a+b\right)^{u}}{3^{u}}$$
Or $a\to a^2$,$b\to b^2$:
$$\left(4-\frac{3\left(ab+b+a\right)}{b^{2}+a^{2}+1}\right)^{\frac{1}{2}}\left(a^{u}+b^{u}+\left(ab\right)^{u}\right)^{\frac{1}{u}}-\frac{3^{\frac{1}{u}}\left(\frac{16}{9\sqrt{3}}\right)^{\frac{1}{6}}\left(1+a^{2}+b^{2}\right)}{3}\leq 0$$
Now we have the inequality :
$$\left(4-\frac{3\left(ab+b+a\right)}{b^{2}+a^{2}+1}\right)^{\frac{1}{2}}\left(a^{u}+b^{u}+\left(ab\right)^{u}\right)^{\frac{1}{u}}-\frac{3^{\frac{1}{u}}\left(\frac{16}{9\sqrt{3}}\right)^{\frac{1}{6}}\left(1+a^{2}+b^{2}\right)}{3}\leq \left(4-\frac{3\left(ab+b+a\right)}{b^{2}+a^{2}+1}\right)^{\frac{1}{2}}\left(a+b+\left(ab\right)\right)-\frac{3\left(\frac{16}{9\sqrt{3}}\right)\left(1+a^{2}+b^{2}\right)}{3}\tag{1}$$
Or $x\in[0,1]$:
$$\left(4-3x\right)^{\frac{1}{2}}x-\frac{16}{9\sqrt{3}}\leq 0$$
Where :
$$x=\frac{ab+b+a}{b^{2}+a^{2}+1}$$
Wich is is obvious .
Some remarks :
I cannot explain or show all the steps notably $(1)$ so how to explain that and have you an other path to show it ?
Almost half an answer .
We have for $0<a=b\leq 1$ and $x\in[0,1]$ :
$$-\frac{\left(\frac{\left(\frac{16}{9\sqrt{3}}\right)^{\frac{1}{6}}\left(1+x^{2}+a^{2}\right)}{3}\right)^{2}}{\left(\frac{x^{u}}{3}+\frac{a^{u}}{3}+\frac{\left(xa\right)^{u}}{3}\right)^{\frac{2}{u}}}\le-\frac{\left(\frac{\left(\frac{16}{9\sqrt{3}}\right)^{\frac{1}{6}}\left(1+x^{2}+b^{2}\right)}{3}\right)^{2}}{\left(\frac{1+2\left(xb\right)^{u}}{3}\right)^{\frac{2}{u}}}\tag{i}$$
So remains to show an inequality with variable $xb=U$ and $x+b=Y$ wich is easier to tackle .We can use $ab1$'s method .
It's not hard using $i$ that the difference for fixed sum $a+b=\alpha$ is maximized for $a=b=\alpha/2$ so we need to show for positive $x$:
$$\left(4-\frac{3\left(x+\frac{x^{2}}{4}\right)}{1+\frac{x^{2}}{2}}\right)\left(\frac{\left(1+2\left(\frac{x^{2}}{4}\right)^{u}\right)}{3}\right)^{\frac{2}{u}}-\frac{\left(\frac{16}{9\sqrt{3}}\right)^{\frac{1}{3}}\left(1+\frac{x^{2}}{2}\right)^{2}}{3^{2}}<0$$
Setting $u\to 0$ we get the inequality :
$$\frac{\left(4-\frac{3\left(x+\frac{x^{2}}{4}\right)}{1+\frac{x^{2}}{2}}\right)x^{\frac{8}{3}}}{4\cdot2^{\frac{2}{3}}}-\frac{\left(\frac{16}{9\sqrt{3}}\right)^{\frac{1}{3}}\left(1+\frac{x^{2}}{2}\right)^{2}}{3^{2}}<0$$
See Wolfram Alpha