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
 A: 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
$$xz+yz-2xyz\le\dfrac{1}{2}-xy\Longleftrightarrow xy+xz+yz\le\dfrac{1}{2}+2xyz$$
A: Edit: I found the following has a false claim that $xyz\ge1/8$ follows from the constraint. I'll leave it up for now (will delete if asked). At the end I give an example where $xyz<1/8.$
Your proof can be finished, given you correct the final inequality. Note the equivalent inequalities $$1-2xyz \le 2xyz+1/2, \\ 1/2 \le 4xyz,\\ xyz\ge 1/8.$$
Now from the constraint $x^2+y^2+z^2+2xyz=1$ and the objective to minimize $xyz$ one can use lagrange multipliers to conclude that, for positive $x,y,z$, one has $x=y=z$ at any critical point. With each of these put equal to $t$ we get $2t^3+3t^2-1=(t+1)^2(2t-1)=0$ giving $t=1/2$ [can't use $t=-1$ here]where $xyz=1/8.$
A bit more work is needed to verify this is indeed the minimum of $xyz$ given the constraint.
Edit: A bit more indeed! It must be these internal critical points don't give the global minimum of $xyz$, since $x=y=3/5,z=7/25$ satisfies the constraint $x^2+y^2+z^2+2xyz=1$ and yet $xyz=63/625=0.1008<1/8=0.125.$
A: 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 $x=(1+\sqrt{17})/8$).
A: Let $x=\frac{a}{\sqrt{(a+b)(a+c)}}$ and $y=\frac{b}{\sqrt{(a+b)(b+c)}}$, where $a$, $b$ and $c$ are positives.
Hence, $z=\frac{c}{\sqrt{(a+c)(b+c)}}$ and we need to prove that
$$\sum_{cyc}\frac{ab}{(a+b)\sqrt{(a+c)(b+c)}}\leq\frac{2abc}{\prod\limits_{cyc}(a+b)}+\frac{1}{2}$$ or
$$2\sum_{cyc}ab\sqrt{(a+c)(b+c)}\leq4abc+\prod_{cyc}(a+b)$$ or
$$2\sum_{cyc}ab\sqrt{(a+c)(b+c)}\leq\sum_{cyc}(a^2b+a^2c+2abc),$$
which is AM-GM:
$$2\sum_{cyc}ab\sqrt{(a+c)(b+c)}\leq\sum_{cyc}ab(a+b+2c)=\sum_{cyc}(a^2b+a^2c+2abc).$$
Done!
