How prove this $xy+yz+xz\le 2xyz+\frac{1}{2}$

We first we first notice that there are always two of the three numbers, both greater than $\dfrac{1}{2}$,because of symmetry,we may assume that $x,y\le\dfrac{1}{2},$ or $x,y\ge \dfrac{1}{2}$ and then $$(2x-1)(2y-1)\ge 0\Longleftrightarrow x+y-2xy\le\dfrac{1}{2}$$ on the other hand, $$1=x^2+y^2+z^2+2xyz\ge 2xy+z^2+2xyz$$ then $$2xy(1+z)\le 1-z^2\Longrightarrow 2xy\le 1-z$$ we only have to multiply side by side the inequality from above $$x+y-2xy\le\dfrac{1}{2},z\le 1-2xy$$ then $$xz+yz-2xyz\le\dfrac{1}{2}-xy\Longleftrightarrow xy+xz+yz\le\dfrac{1}{2}+2xyz$$