Edit(20230517)
WLOG, assume that $c = \min(a,b,c)$.
Since $x\mapsto \sqrt[3]{x}$ is concave on $(0, \infty)$, we have
\begin{align*}
\sqrt[3]{1+2ac} + \sqrt[3]{1+2ba} + \sqrt[3]{1+2cb}
&\le 2\sqrt[3]{\frac{1+2ac + 1+2cb}{2}} + \sqrt[3]{1+2ba}\\
&= 2\sqrt[3]{1+ac + cb} + \sqrt[3]{1+2ba}. \tag{1}
\end{align*}
It suffices to prove that
$$2\sqrt[3]{1+ac + cb} + \sqrt[3]{1+2ba} \le 3\sqrt[3]{3}. \tag{2}$$
We split into two cases.
Case 1: $ba < 1$
Using $c = \min(a, b, c)$, we have $ac + cb \le 2ba\le 2$. Thus, (2) is true.
Case 2: $ba \ge 1$
From $a+b+c+abc = 4$, we have
$c = \frac{4-a-b}{ab+1}$. Also, we have $a+b \ge 2\sqrt{ab} > 2$.
Thus, we have $$ac + cb = \frac{(4-a-b)(a+b)}{ab + 1}
= \frac{4 - (a+b - 2)^2}{ab+1} \le \frac{4 - (2\sqrt{ab} - 2)^2}{ab+1}. \tag{3}$$
Thus, it suffices to prove that
$$2\sqrt[3]{1+ \frac{4 - (2\sqrt{ab} - 2)^2}{ab+1}} + \sqrt[3]{1+2ba} \le 3\sqrt[3]{3}.$$
Let $x = \sqrt{ba}$. Then $1 \le x \le 2$. It suffices to prove that, for all $1\le x\le 2$,
$$2\sqrt[3]{1+ \frac{4 - (2x - 2)^2}{x^2+1}} + \sqrt[3]{1+2x^2} \le 3\sqrt[3]{3}$$
or
$$2\sqrt[3]{1 - \frac{2(3x-1)(x-1)}{3(x^2+1)}} + \sqrt[3]{\frac{1+2x^2}{3}} \le 3. \tag{4}$$
Let
$$A := 1 - \frac{2(3x-1)(x-1)}{3(x^2+1)}, \quad B := \frac{1+2x^2}{3}.$$
Then $0 < A \le 1$ and $B \ge 1$.
(4) is written as
$$\sqrt[3]{B} - 1 \le 2(1 - \sqrt[3]{A})$$
or
$$\frac{B - 1}{B^{2/3} + B^{1/3} + 1}
\le \frac{2(1 - A)}{A^{2/3} + A^{1/3} + 1}.$$
We have $B^{2/3} + B^{1/3} \ge 2\sqrt{B^{2/3} \cdot B^{1/3}} = 2\sqrt{B} = \frac{4B}{2\sqrt{B}} \ge \frac{4B}{B + 1}$.
Also, we have $A^{2/3} + A^{1/3} \le 2A^{1/3}
\le 2\cdot \frac{A + 1 + 1}{3} = \frac{2A + 4}{3}$.
Thus, it suffices to prove that
$$\frac{B - 1}{\frac{4B}{B+1} + 1}
\le \frac{2(1 - A)}{\frac{2A + 4}{3} + 1}$$
or
$$\frac{2(x-1)(x+1)(x^2+2)}{15x^2 + 12} \le \frac{12(3x-1)(x-1)}{15x^2 + 16x + 23}$$
or
$$\frac{2(x+1)(x^2+2)}{15x^2 + 12} \le \frac{12(3x-1)}{15x^2 + 16x + 23}$$
or (after clearing the denominators)
$$-15\,{x}^{4}-46\,{x}^{3}+155\,{x}^{2}-20\,x+118 \ge 0$$
which is true. Indeed, we have
$$\mathrm{LHS} \ge -15x^2 \cdot 2^2 - 46x^2 \cdot 2 + 155x^2 - 20 \cdot 2 + 118 = 3x^2 + 78 > 0.$$
We are done.
Some thoughts.
By Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align*}
&\left(\frac{a}{\sqrt{8a+bc}}+\frac{b}{\sqrt{8b+ca}}+\frac{c}{\sqrt{8c+ab}}\right)^2\\[6pt]
\le{}& \left(\sum_{\mathrm{cyc}}\frac{a^2}{(8a + bc)(3a +abc )}\right)\left(\sum_{\mathrm{cyc}} (3a + abc)\right)\\[6pt]
={}& \left(\sum_{\mathrm{cyc}}\frac{a^2}{(8a + bc)(3a +abc )}\right)\cdot 3(a + b + c + abc).
\end{align*}
It suffices to prove that
$$\sum_{\mathrm{cyc}}\frac{a^2}{(8a + bc)(3a +abc )} \le \frac{1}{12}. \tag{1}$$
(1) is true which is verified by Mathematica.
Best Answer
Let $a=\frac{8kyz}{x^2},$ $b=\frac{8kxz}{y^2}$ and $c=\frac{8kxy}{z^2},$ where $x$, $y$, $z$ and $k$ are positive numbers.
Thus, the condition gives $k\geq1$ and we need to prove that $$\sum_{cyc}\frac{x}{\sqrt{x^2+8xyz}}\geq\frac{3}{\sqrt{1+8k}}.$$ Now, by Holder $$\left(\sum_{cyc}\frac{x}{\sqrt{x^2+8xyz}}\right)^2\sum_{cyc}x(x^2+8kyz)\geq(x+y+z)^3.$$ Id est, it's enough to prove that $$(1+8k)(x+y+z)^3\geq9\sum_{cyc}x(x^2+8kyz)$$ or $$\sum_{cyc}(8(k-1)x^3+3(1+8k)(x^2y+x^2z)-(56k-2)xyz)\geq0$$ or $$8(k-1)\sum_{cyc}(x^3-x^2y-x^2z+xyz)+(32k-5)\sum_{cyc}z(x-y)^2\geq0,$$ which is true by Schur.