<s>[INST] With $xy+yz+zx=-1$, proving: $x^2+2y^2+2z^2 .....$ Assuming $xy+yz+zx=-1$, prove that :

$$x^2+2y^2+2z^2 \geq \frac{1+\sqrt{17}}{2}$$

 [/INST] We need to prove that
$$x^2+2y^2+2z^2+\frac{1+\sqrt{17}}{2}(xy+xz+yz)\geq0$$ or
$$x^2+\frac{1+\sqrt{17}}{2}(y+z)x+2y^2+2z^2+\frac{1+\sqrt{17}}{2}yz\geq0,$$
for which it's enough to prove that
$$\left(\frac{1+\sqrt{17}}{2}(y+z)\right)^2-4\left(2y^2+2z^2+\frac{1+\sqrt{17}}{2}yz\right)\leq0,$$
which is $(y-z)^2\geq0$.
Done!
</s>