Prove inequality $(x+y+z-2xyz)^2 \le 2$ 
Problem: Prove inequality $(x+y+z-2xyz)^2  \le 2\ (1)$ with
  $x^2+y^2+z^2 = 1 \land x,y,z \in \mathbb R$

I tried expand $LHS$ and have:
$$(1)\iff 1 - 2 (xy+yz+xz) + 4 xyz(x+y+z)-(2xyz)^2 \ge 0$$
Denote: $xy = a, yz = b,xz=c \implies (1) \iff1-2\sum a+ 4 \sum ab - 2abc \ge0$
But stuck. Please help me.

we let $x\le y\le z$,then $$z^2\ge\dfrac{1}{3},2xy\le x^2+y^2\le\dfrac{2}{3}$$
use Cauchy-Schwarz inequality we have
$$(x+y+z-2xyz)^2=[(x+y)+z(1-2xy)]^2\le [(x+y)^2+z^2][1+(1-2xy)^2]$$
we only prove
$$[(x+y)^2+z^2][1+(1-2xy)^2]\le 2$$
since:
$$ [(x+y)^2+z^2][1+(1-2xy)^2]=(1+2xy)[1+(1-2xy)^2]=2+4x^2y^2(2xy-1)\le 2$$
By done!
