<s>[INST] 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.
 [/INST] Expanding the $LHS$ we have 
\begin{align}1+ 2(xy+ xz+ yz) - 4xyz(x+y+z) + 4x^2y^2z^2 \leq 3- 4xyz(x+y+z) + 4x^2y^2z^2
\end{align}
since $(xy+ xz+ yz) \leq (x^2+y^2+z^2)=1$ by Cauchy-Schwarz inequality. Then
\begin{align} 3- 4xyz(x+y+z) + 4x^2y^2z^2 &= 3- 4x^2y^2z^2(1/xy+1/yz+1/xz) + 4x^2y^2z^2\\& \leq 3- 32 x^2y^2z^2
\end{align}
since $\frac{3}{(1/xy+1/yz+1/xz)}\leq \frac{(x^2+y^2+z^2)}{3}\leq \frac{1}{3}$ by harmonic and arithmetic means inequality and the previous estimation.
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