With the notation $d=\frac{1}{a b c}$ the inequality becomes
$$\frac{1}{a+b+d+1}+\frac{1}{b+c+d+1}+\frac{1}{c+a+d+1}+\frac{1}{a+b+c+1} \leq 1 .$$
Let $x=\sqrt[4]{a}$, $y=\sqrt[4]{b}$, $z=\sqrt[4]{c}$, $w=\sqrt[4]{d}$. Then
$$x y z w=\sqrt[4]{a b c d}=1$$
and it suffices to prove that
$$\begin{equation*} \sum \frac{1}{x^{4}+y^{4}+z^{4}+z y x w} \leq 1 . \tag{1} \end{equation*}$$
But $x^{4}+y^{4}+z^{4} \geq x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}$ so
$$\begin{aligned} x^{4}+y^{4}+z^{4} & =\frac{x^{4}+y^{4}+z^{4}+x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}{2} \\ & =\frac{x^{4}+y^{2} z^{2}}{2}+\frac{y^{4}+z^{2} x^{2}}{2}+\frac{z^{4}+x^{2} y^{2}}{2} \\ & \geq x^{2} y z+y^{2} z x+z^{2} x y=x y z(x+y+z) \end{aligned}$$
where we used the AM-GM inequality three times. The same result follows from the rearrangement inequality. Hence
$$\begin{gathered} \frac{1}{x^{4}+y^{4}+z^{4}+x y z w} \leq \frac{1}{x y z(x+y+z)+x y z w} \\ =\frac{1}{x y z(x+y+z+w)}=\frac{w}{x+y+z+w} \end{gathered}$$
and so
$$\sum \frac{1}{x^{4}+y^{4}+z^{4}+x y z w} \leq \sum \frac{w}{x+y+z+w}=1$$