Given inequality is equivalent to:
$$\frac{1}{1+\frac{x}{y}+\frac{y}{x}}+\frac{1}{1+\frac{y}{z}+\frac{z}{y}}+\frac{1}{1+\frac{z}{x}+\frac{x}{z}} \le \frac{1}{2+\frac{z}{x}}+\frac{1}{2+\frac{x}{y}}+\frac{1}{2+\frac{y}{z}}.$$
Now we can take substitution $\frac{x}{y} = a$, $\frac{y}{z} = b$, $\frac{z}{x} = c \Rightarrow abc = 1$, so our inequality becomes:

$$\frac{1}{1+a+\frac{1}{a}}+\frac{1}{1+b+\frac{1}{b}}+\frac{1}{1+c+\frac{1}{c}} \le \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}.$$
After some computations and also using $abc = 1$, this becomes equivalent to:
$$\frac{3(a+b+c)+3(ab+bc+ca)+(ab+bc+ca)(a+b+c)}{(ab+bc+ca)^2+(ab+bc+ca)(a+b+c)+(a+b+c)^2} \le \frac{12+4(a+b+c)+(ab+bc+ca)}{9+2(ab+bc+ca)+4(a+b+c)}.$$
Now, we take substitution $a+b+c = S$, $ab+bc+ca = P$ and inequality becomes:
$$\frac{3S+3P+SP}{P^2+PS+S^2} \le \frac{12+4S+P}{9+2P+4S} \Leftrightarrow P^3+4S^3+3P^2S+PS^2+6P^2 \ge 27S+27P+15PS \quad (*)$$
It is not difficult to prove that $S^2 \ge 3P$ and also $S \ge 3, P \ge 3$ (by AM $\ge$ GM and $abc = 1$). Therefore:
$$\begin{aligned} 4S^3 &= 4S^2 \cdot S \ge 12PS, \quad PS^2 = S \cdot PS \ge 3PS, \quad 3P^2S \ge 3 \cdot 3^2 \cdot S = 27S, \\ P^3 &= P^2 \cdot P \ge 3^2 \cdot P = 9P, \quad 6P^2 = 6P \cdot P \ge 6 \cdot 3P = 18P. \end{aligned}$$
By summing these inequalities we found (*) to be true so our proof is finished. Equality is obviously achieved when
$$S^2 = 3P = 9 \Rightarrow a = b = c = 1 \Rightarrow x = y = z.$$