A squared cubic sum bounds three quadratic factors

Let $ab=z^2$ , $ac=y^2$ and $bc=x^2$ , where $x$ , $y$ and $z$ are positive numbers. Hence, we need to prove that $(x^3+y^3+z^3+xyz)^2\geq2(x^2+y^2)(x^2+z^2)(y^2+z^2)$ , which is $\sum_{cyc}(x^6-2x^4y^2-x^4z^2+2x^3y^3+2x^4yz-x^2y^2z^2)\geq0$ , which is obvious (Schur and SOS).

Source: InternLM Lean-Workbook, record lean_workbook_6019 (Apache-2.0). Complete source proposition preserved; proof developed independently.