With $xy+yz+zx=-1$, proving: $x^2+2y^2+2z^2 .....$ Assuming $xy+yz+zx=-1$, prove that :

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

 A: You can do this by suitable coefficients and a variant of Cauchy-Schwarz inequality.  
First, note that $\dfrac{a_1^2}{b_1} + \dfrac{a_2^2}{b_2} + \dfrac{a_3^2}{b_3} \ge \dfrac{(a_1 + a_2 + a_3)^2}{b_1 + b_2+b_3}$.  This readily follows from Cauchy Schwarz, for real numbers $a_i$ and $b_i$.  
So, $\dfrac{x^2}{a} + \dfrac{y^2}{b} + \dfrac{z^2}{b} \ge \dfrac{(x + y + z)^2}{a + 2b} = \dfrac{x^2 + y^2 + z^2 - 2}{a + 2b}$
$$ \implies \left( \frac{1}{a} - \frac{1}{a+2b} \right)x^2 + \left( \frac{1}{b} - \frac{1}{a+2b} \right)\left(y^2+z^2\right) \ge \frac{-2}{a+2b}$$
Comparing with the LHS we want in our equality, we have two equations to solve to get the right $a, b$ :
$$ \frac{1}{a} - \frac{1}{a+2b} = 1 \text{ and } \frac{1}{b} - \frac{1}{a+2b} = 2$$
Solving these, you have two possible solutions, choose the one which gives you the desired  RHS.  
In this case, $b = \dfrac{7+\sqrt{17}}{8} $ and $a = -\dfrac{3+\sqrt{17}}{2} $ do the trick and give RHS of $\dfrac{1+\sqrt{17}}{2} $
A: 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!
