<s>[INST] Smallest value of $x^2+5y^2+8z^2$ given $xy+yz+xz=-1$. The question is:

Find the smallest value of $x^2+5y^2+8z^2$ given $xy+yz+xz=-1$.

Here's what I've tried so far. Dividing by $xyz$, I get $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{-xyz}$. Squaring both sizes:
$$\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)^2 = \frac{1}{(xyz)^2}$$
I tried applying Cauchy Schwartz on this:
$$\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)^2 \leq 3 \left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)$$
but after this I'm stuck. Can someone help me out here?
I have also tried AM-GM on $x^2+5y^2+8z^2$ but that doesn't seem to lead anywhere.
Wolfram alpha says the answer is $4$.
 [/INST] Let $k$ be a minimal value.
Thus, $k>0$ and $$x^2+5y^2+8z^2\geq k$$ or
$$x^2+5y^2+8z^2+k(xy+xz+yz)\geq0$$ or
$$x^2+k(y+z)x+5y^2+8z^2+kyz\geq0,$$ for which we need 
$$k^2(y+z)^2-4(5y^2+8z^2+kyz)\leq0$$ or
$$(20-k^2)y^2+(4k-2k^2)yz+(32-k^2)z^2\geq0,$$ for which we need
$$20-k^2>0$$ and $$(2k-k^2)^2-(20-k^2)(32-k^2)=0$$ or
$$k^3-14k^2+160=0$$ or
$$k^3-4k^2-10k^2+40k-40k+160=0$$ or
$$(k-4)(k^2-10k-40)=0,$$ which gives $k=4$.
Easy to see that the equality occurs, which says that we got a minimal value.
Now, we see that
$$x^2+5y^2+8z^2\geq4$$ it's
$$x^2+4(y+z)x+5y^2+8z^2+4yz\geq0$$ or
$$(x+2y+2z)^2+(y-2z)^2\geq0.$$
The equality occurs for $y=2z$, $x=-6z$ and $xy+xz+yz=-1.$
Id est, for example, for  $$(x,y,z)=\left(-\frac{3}{2}, \frac{1}{2}, \frac{1}{4}\right).$$
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