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

Here's a solution using the Lagrange multiplier method. We are looking for extrema of $f(x,y,z) = xy+yz+xz-2xyz$ given that $x,y,z \geq 0$ and $x^2+y^2+z^2+2xyz=1$. On the boundary (where one variable is $0$) we have $f \in [0,\tfrac{1}{2}]$. Now assume $x,y,z>0$. Define $s=x+y+z$.  Then the Lagrange multiplier method shows that at an extremum of $f$ we must have
$$\begin{eqnarray}
(y-z)(x^2-2sx+s) &=& 0\\
(x-z)(y^2-2sy+s) &=& 0\\
(x-y)(z^2-2sz+s) &=& 0
\end{eqnarray}
$$
Consider the first equality.  This implies $y=z$ or $x > \tfrac{1}{2}$.  Since not all of $x,y,z$ can be greater than $\tfrac{1}{2}$ we conclude that at least two of these variables must be equal.  Now $f(x,x,1-2x^2)$ has a maximum of $\tfrac{1}{2}$ on the interval $(0,\sqrt{1/2})$ at $x=\tfrac{1}{2}$ (and a local minimum at $x=(1+\sqrt{17})/8$).
