Prove this inequality with $xyz\le 1$

Solution 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.