Prove this inequality with $xyz\le 1$ 
If $x,y,z\ge 0$ and $\color{red}{xyz\le 1}$, show that
$$\color{blue}{\dfrac{x^2-x+1}{x^2+y^2+1}+\dfrac{y^2-y+1}{y^2+z^2+1}
+\dfrac{z^2-z+1}{z^2+x^2+1}\ge 1}.$$


This inequality admits an SOS (Sum of Squares) representation
\begin{align}
&\frac{x^2-x+1}{x^2+y^2+1} + \frac{y^2-y+1}{y^2+z^2+1} + \frac{z^2-z+1}{z^2+x^2+1} - 1 \\
=\ & \frac{1}{(x^2+y^2+1)(y^2+z^2+1)(z^2+x^2+1)}\Big[(1-xyz)(f_1^2+f_2^2+f_3^2) + \frac{1}{12}u^TQu\Big]
\end{align}
where $f_1, f_2, f_3$ are polynomials with integer coefficients,
$u = [1, x, y, z, xy, xz, yz, x^2, y^2, z^2, xyz, x^2y, xz^2, y^2z]^T$
and $Q$ is a positive semi-definite (PSD) $14\times 14$ constant integer matrix.
The desired result follows.
Remarks: 
1) Since the problem is tagged as contest-math but my proof is not an human proof, I do not give $f_1, f_2, f_3$ and $Q$ currently
and hope to see nice proofs.
2) Mathematica Resolve can prove the inequality although we do not see the step-by-step solution.
