Remarks: Since no verified proof is available, I give a proof. It is a computer assisted proof. It is very complicated by hand. Hope to see nice proofs.
Sketch of a proof.
Let
$$x := \frac{a + b}{c + ab},
\quad y := \frac{b + c}{a + bc},
\quad z := \frac{c + a}{b + ca},$$
$$u := \frac{2(a + b)}{ab + a + b + c},
\quad v := \frac{2(b + c)}{bc + a + b + c}, \quad w := \frac{2(c + a)}{ca + a + b + c}.$$
By AM-GM, we have $u = \frac{2x}{1 + x} \le \sqrt x$, $v = \frac{2y}{1+y} \le \sqrt{y}$,
and $w = \frac{2z}{1+z}\le \sqrt{z}$.
We have
$$\sqrt{x} - u = \frac{x - u^2}{\sqrt{x} + u} \ge \frac{x - u^2}{\frac{x + 1}{2} + u}. \tag{1}$$
Using (1), it suffices to prove that
$$u + \frac{x - u^2}{\frac{x + 1}{2} + u} + v + \frac{y - v^2}{\frac{y + 1}{2} + v} + w + \frac{z - w^2}{\frac{z + 1}{2} + w} \ge 3. \tag{2}$$
We use pqr method.
Let $p = a + b + c, q = ab + bc + ca, r = abc$.
The condition $a + b + c + abc = 4$ is written as
$$p + r = 4. \tag{3}$$
(2) is written as
\begin{align*}
&5\,{p}^{6}+10\,{p}^{5}q-62\,{p}^{5}r+{p}^{4}{q}^{2}-14\,{p}^{4}qr-3\,{
p}^{4}{r}^{2}+24\,{p}^{4}q\\
&\quad +144\,{p}^{4}r-16\,{p}^{3}{q}^{2}+48\,{p}^{3
}qr-126\,{p}^{3}{r}^{2}-8\,{p}^{2}{q}^{3}-8\,{p}^{2}{q}^{2}r\\
&\quad -30\,{p}^{
2}q{r}^{2}-6\,{p}^{2}{r}^{3}-96\,{p}^{3}r+16\,{p}^{2}{q}^{2}+288\,{p}^
{2}qr+252\,{p}^{2}{r}^{2}\\
&\quad -224\,p{q}^{3}+384\,p{q}^{2}r+72\,pq{r}^{2}-
64\,p{r}^{3}-48\,{q}^{4}+64\,{q}^{3}r+8\,{q}^{2}{r}^{2}\\
&\quad -16\,q{r}^{3}-3
\,{r}^{4}-256\,pqr-352\,p{r}^{2}+128\,{q}^{2}r-320\,q{r}^{2}+96\,{r}^{
3}+192\,{r}^{2}\\
&\ge 0. \tag{4}
\end{align*}
From (3), using $r = 4 - p$, (4) is written as
\begin{align*}
&-48\,{q}^{4}+ \left( -8\,{p}^{2}-288\,p+256 \right) {q}^{3}+ \left( {p
}^{4}-8\,{p}^{3}-392\,{p}^{2}+1344\,p+640 \right) {q}^{2}\\
&\quad + \left( 24\,
{p}^{5}-110\,{p}^{4}+232\,{p}^{3}-160\,{p}^{2}+3456\,p-6144 \right) q \\
&\quad +
64\,{p}^{6}-488\,{p}^{5}+1873\,{p}^{4}-5296\,{p}^{3}+10592\,{p}^{2}-
15104\,p+8448\\
&\ge 0. \tag{5}
\end{align*}
From (3), we have $3 \le p \le 4$.
Using $q^2 \ge 3pr$ and (3),
we have $q \ge \sqrt{3p(4 - p)}$.
Using degree four Schur inequality $r \ge \frac{5p^2q - p^4 - 4q^2}{6p}$ and (3), we have
$$4 - p \ge \frac{5p^2q - p^4 - 4q^2}{6p}$$
which results in (using $p^2 \ge 3q$)
$$q \le \frac58 p^2 - \frac18\sqrt{9p^4 + 96p^2 - 384p}.$$
Thus, we have
$$3 \le p \le 4, \quad \sqrt{3p(4 - p)} \le q \le \frac58 p^2 - \frac18\sqrt{9p^4 + 96p^2 - 384p}. \tag{6}$$
If $q = \sqrt{3p(4 - p)}$, we can prove (5).
If $\sqrt{3p(4 - p)} < q \le \frac58 p^2 - \frac18\sqrt{9p^4 + 96p^2 - 384p}$, we use the Buffalo Way (BW).
Let
$$q_1 := \sqrt{3p(4 - p)}, \quad q_2 := \frac58 p^2 - \frac18\sqrt{9p^4 + 96p^2 - 384p}.$$
Let
$$s := \frac{q_2 - q}{q - q_1} \ge 0.$$
We have
$$q = q_1 \cdot \frac{s}{1 + s} + q_2 \cdot \frac{1}{1 + s}. \tag{7}$$
We plug (7) into (5), after clearing the denominators, it suffices to prove that
$$m_4s^4 + m_3s^3 + m_2s^2 + m_1s + m_0 \ge 0 \tag{8}$$
where $m_0, m_1, m_2, m_3, m_4$ are functions in $p$.
We can prove that $m_0, m_1, m_2, m_3, m_4 \ge 0$ for all $3\le p\le 4$. Thus, (8) is true.
We are done.