How prove this $xy+yz+xz\le 2xyz+\frac{1}{2}$ let $x,y,z>0$ and such
$$x^2+y^2+z^2+2xyz=1$$
show that
$$xy+yz+xz\le 2xyz+\dfrac{1}{2}$$
My try: since
$$1=x^2+y^2+z^2+2xyz\ge xy+yz+xz+2xyz$$
then
$$xy+yz+xz\le 1-2xyz$$
so
we only prove follow this 
$$1-2xyz\le 2xyz+\dfrac{1}{2}$$
$$\Longleftrightarrow xyz\ge\dfrac{1}{4}$$
But this is not true,so How prove my inequality? Thank you

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
$$xz+yz-2xyz\le\dfrac{1}{2}-xy\Longleftrightarrow xy+xz+yz\le\dfrac{1}{2}+2xyz$$
