How prove this $x^2+y^2+z^2+3\ge 2(xy+yz+xz)$ let $x,y,z$ be real numbers, and such $$x+y+z+xyz=4$$ show that
$$x^2+y^2+z^2+3\ge 2(xy+yz+xz)$$
My try: let
$$x+y+z=p,xy+yz+xz=q,xyz=r$$
then
$$p+r=4$$
then
$$\Longleftrightarrow p^2+3\ge 4q$$
But I can't.Thank you
 A: Note that it is sufficient to consider $x, y, z \ge 0$.  Then the condition yields $4 = x+y+z+xyz \ge 3\sqrt[3]{xyz}+xyz \implies xyz \le 1$.
Also note that among $x-1, y-1, z-1$, at least two have the same sign.  WLOG let $(y-1)(z-1) \ge 0$.  Then we have:
$$(x-1)^2+(y-z)^2+2x(y-1)(z-1) \ge 0$$
$$\implies x^2+y^2+z^2+2xyz + 1 \ge  2(xy+yz+zx)$$
which along with $xyz \le 1$ gives you the result.
A: I think you should consider $f(x,y,z): {\bf R}^3 \to {\bf R}$ defined as 
$$(x,y,z) \mapsto x^2 + y^2 + z^2 -2xy -2yz -2xz + 3.$$
Then $f \in {\mathcal C}^\infty({\bf R}^3)$ and you could investigate where $f(x,y,z) = 0$. That could get you a little further.
A: Following the suggestion by TooOldForMath given in the comments above we may define
$$
f(x,y,z)=x^2+y^2+z^2+3-2(xy+yz+xz)
$$
which is the multivariate function that we want to show has minimal value greater than or equal to zero when subject to the constraint
$$
g(x,y,z)=x+y+z+xyz=4
$$
so we define
$$
\Lambda(x,y,z,\lambda)=f(x,y,z)+\lambda(g(x,y,z)-4)
$$
and calculate the corresponding system of equations
$$
\begin{align}
\frac{\partial\Lambda}{\partial x}&=2x-2y-2z+\lambda(1+yz)&=0\\
\frac{\partial\Lambda}{\partial y}&=2y-2x-2z+\lambda(1+xz)&=0\\
\frac{\partial\Lambda}{\partial z}&=2z-2x-2y+\lambda(1+xy)&=0\\
\frac{\partial\Lambda}{\partial \lambda}&=x+y+z+xyz-4&=0
\end{align}
$$
This system is symmetrical in $x,y$ and $z$ and can be solved by Gauss elimintation to get the solutions
$$
x=y=z=\lambda=1\\
\mbox{and}\\
x=4,y=z=-0.5,\lambda=-8
$$
In the last solution one may interchange $x,y$ and $z$ to get three essentially identical solutions. Moreover there is a complex valued solution too, but that is of no relevance to the current question. Plugging in $(x,y,z)=(1,1,1)$ and $(x,y,z)=(4,-0.5,-0.5)$ respectively we obtain the minimum and maximum of the function, namely
$$
\begin{align}
f(1,1,1)&=0\\
f(4,-0.5,-0.5)&=27
\end{align}
$$
Proving that $f$ indeed is greater than or equal to zero everywhere on the curve defined by $g(x,y,z)=4$. In fact $g(x,y,z)=4\implies f(x,y,z)\in[0,27]$. The claim follows!
