This becomes quite easy by just substituting: Writing $x=\frac{a^2}{bc-ab-ac}$ and $y=\frac{b^2}{ac-ba-bc}$ and $z=\frac{c^2}{ab-ca-cb}$ one just checks that \[xy+xz+yz+2xyz=1.\]So if $\vert x\vert+\vert y\vert+\vert z\vert<\frac{3}{2}$ then by mean inequalities also \[\vert xy+xz+yz\vert \le \vert x\vert \cdot \vert y\vert+\vert x\vert \cdot \vert z\vert+\vert y\vert \cdot \vert z\vert <\frac{3}{4}\]and \[\vert xyz\vert<\frac{1}{8}\]which is a fair contradiction.