another inequality $\frac{1}{xy+z}+\frac{1}{yz+x}+\frac{1}{zx+y}\le\frac{1}{2}$ 
Let $x,y,z>0$ and such $xyz\ge 2+x+y+z$, show that
  $$\dfrac{1}{xy+z}+\dfrac{1}{yz+x}+\dfrac{1}{zx+y}\le\dfrac{1}{2}\tag{1}$$

I have posted this problem. Unfortunately, it doesn't hold, but I guess $(1)$ may hold.
The theory basis of speculation Use the following classical results
$$xyz=2+x+y+z\Longrightarrow xy+yz+zx\ge 2(x+y+z)$$
 A: I agree with you. Your inequality is true!
Indeed, the condition gives $\sum\limits_{cyc}\frac{1}{x+1}\leq1$.
Let $x=a$, $y=b$ and $z=kc$, where $k>0$ and $\sum\limits_{cyc}\frac{1}{a+1}=1$. Hence,
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{kc+1}$$
which gives $k\geq1$.
Thus, $\sum\limits_{cyc}\frac{1}{xy+z}=\frac{1}{ab+kc}+\frac{1}{kac+b}+\frac{1}{kbc+a}\leq\frac{1}{ab+c}+\frac{1}{ac+b}+\frac{1}{bc+a}$.
Id est, it remains to prove that $\sum\limits_{cyc}\frac{1}{ab+c}\leq\frac{1}{2}$ for positives $a$, $b$ and $c$ such that $\sum\limits_{cyc}\frac{1}{a+1}=1$.
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 $\sum\limits_{cyc}\frac{1}{\frac{x+y}{z}\cdot\frac{x+z}{y}+\frac{y+z}{x}}\leq\frac{1}{2}$ or
$$\sum\limits_{cyc}\frac{xyz}{x(x+y)(x+z)+yz(y+z)}\leq\frac{1}{2}$$
or
$$\sum\limits_{cyc}\frac{xyz}{(x+y+z)(x^2+yz)}\leq\frac{1}{2}$$
or
$$\sum\limits_{cyc}\left(\frac{x}{2}-\frac{xyz}{x^2+yz}\right)\geq0$$
or
$$\sum\limits_{cyc}\frac{x(x^2-yz)}{x^2+yz}\geq0$$
or
$$\sum\limits_{cyc}\frac{x((x-y)(x+z)-(z-x)(x+y))}{x^2+yz}\geq0$$
or
$$\sum\limits_{cyc}(x-y)\left(\frac{x(x+z)}{x^2+yz}-\frac{y(y+z)}{y^2+xz}\right)\geq0$$
or
$$\sum\limits_{cyc}z(x-y)^2(x^2+y^2+xz+yz)(z^2+xy)\geq0$$
Done!
