Should it be $x^2+y^2+z^2+xy+yz+zx \le 1$ ?

parmenides51 wrote: Positive numbers $x, y, z$ satisfy $x^2+y^2+z^2+xy+yz+zx \le 1$. Prove that $\big( \frac{1}{x}-1\big) \big( \frac{1}{y}-1\big)\big( \frac{1}{z}-1\big) \ge 9 \sqrt6 -19$. We need to prove that $$\prod_{cyc}\left(\frac{\sqrt{\sum\limits_{cyc}(x^2+xy)}}{x}-1\right)\geq9\sqrt6-19,$$which is a linear inequality of $w^3$. We even can say more. It's equivalent to $f(w^3)\geq0,$ where $f$ decreases. Thus, it's enough to prove the last inequality for the maximal value of $w^3$, which happens for an equality case of two variables. Since the last inequality is homogeneous already, we can assume that $y=z=1$, which gives $$\left(\sqrt{x^2+2x+3}-x\right)\left(\sqrt{x^2+2x+3}-1\right)^2\geq(9\sqrt6-19)x$$or $$(x+2)^4(x^2+2x+3)\geq\left(x^3+4x^2+(9\sqrt6-11)x+6\right)^2$$or $$(x-1)^2(2x^3+(53-18\sqrt6)x^2+(284-108\sqrt6)x+12)\geq0,$$which is obvious.

Positive numbers $x, y, z$ satisfy $x^2+y^2+z^2+xy+yz+zy \le 1$. Prove that $$\big( \frac{1}{x^2}-1\big) \big( \frac{1}{y^2}-1\big)\big( \frac{1}{z^2}-1\big) \ge 125$$$$\big( \frac{1}{x}-x\big) \big( \frac{1}{y}-y\big)\big( \frac{1}{z}-z\big) \ge \frac{125\sqrt6}{36}$$$$\big( \frac{1}{x}-x^2\big) \big( \frac{1}{y}-y^2\big)\big( \frac{1}{z}-z^2\big) \ge \frac{73\sqrt6}{12}-\frac{649}{216}$$