another inequality 1 x y + z + 1 y z + x + 1...

I agree with you. Your inequality is true!
Indeed, the condition gives $\sum _{cyc}\frac{1}{x+1}\le 1$
Let $x=a$, $y=b$ and $z=kc$, where $k>0$ and $\sum _{cyc}\frac{1}{a+1}=1$. Hence,
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{kc+1}$
which gives $k\ge 1$
Thus, $\sum _{cyc}\frac{1}{xy+z}=\frac{1}{ab+kc}+\frac{1}{kac+b}+\frac{1}{kbc+a}\le \frac{1}{ab+c}+\frac{1}{ac+b}+\frac{1}{bc+a}$
Id est, it remains to prove that $\sum _{cyc}\frac{1}{ab+c}\le \frac{1}{2}$ for positives $a$, $b$ and $csuch\; that$ \sum _{cyc}\frac{1}{a+1}=1$$
Let $a=\frac{y+z}{x}$ and $b=\frac{x+z}{y}$, where $x$, $y$ and $z$ are positive numbers.
Hence, $c=\frac{x+y}{z}$ and we need to prove that $\sum _{cyc}\frac{1}{\frac{x+y}{z}\cdot \frac{x+z}{y}+\frac{y+z}{x}}\le \frac{1}{2}$ or
$\sum _{cyc}\frac{xyz}{x(x+y)(x+z)+yz(y+z)}\le \frac{1}{2}$
or
$\sum _{cyc}\frac{xyz}{(x+y+z)(x^2+yz)}\le \frac{1}{2}$
or
$\sum _{cyc}(\frac{x}{2}-\frac{xyz}{x^2+yz})\ge 0$
or
$\sum _{cyc}\frac{x(x^2-yz)}{x^2+yz}\ge 0$
or
$\sum _{cyc}\frac{x((x-y)(x+z)-(z-x)(x+y))}{x^2+yz}\ge 0$
or
$\sum _{cyc}(x-y)(\frac{x(x+z)}{x^2+yz}-\frac{y(y+z)}{y^2+xz})\ge 0$
or
$\sum _{cyc}z(x-y{)}^2(x^2+y^2+xz+yz)(z^2+xy)\ge 0$
Done!